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32. Longest Valid Parentheses
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<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br></pre></td><td class="code"><pre><span class="line">public class Solution {</span><br><span class="line"> private int[] stack = new int[100000];</span><br><span class="line"> private int top = -1;</span><br><span class="line"> private boolean empty = true;</span><br><span class="line"> </span><br><span class="line"> private void push(int x){</span><br><span class="line"> stack[++top] = x;</span><br><span class="line"> if(empty = true) {</span><br><span class="line"> empty = false;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> private void pop(){</span><br><span class="line"> top--;</span><br><span class="line"> if(top == -1){</span><br><span class="line"> empty = true;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> public int longestValidParentheses(String s) {</span><br><span class="line"> char[] str = s.toCharArray();</span><br><span class="line"> int length = str.length;</span><br><span class="line"> int max_length = 0;</span><br><span class="line"> int last_locate = -1;</span><br><span class="line"> for(int i = 0; i < length; i++) {</span><br><span class="line"> if(str[i] == '(') {</span><br><span class="line"> push(i);</span><br><span class="line"> }</span><br><span class="line"> else {</span><br><span class="line"> if(empty) {</span><br><span class="line"> last_locate = i;</span><br><span class="line"> continue;</span><br><span class="line"> }</span><br><span class="line"> else {</span><br><span class="line"> pop();</span><br><span class="line"> if(empty) {</span><br><span class="line"> max_length = max_length > i - last_locate ? max_length : i - last_locate;</span><br><span class="line"> }</span><br><span class="line"> else {</span><br><span class="line"> max_length = max_length > i - stack[top] ? max_length : i - stack[top];</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return max_length;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>这个题目没做出来，参考了别人答案。<br>最初思路的错误在于：<br>1、以为和20题一样，只考虑是否匹配，但是毕竟是Difficulty: Hard<br>2、关键在于是，如何记录当前合法的括号串长度，<br>对于”()((()”这种串，以下两种<br> （1）入栈长度+1,出栈+1，右括弧不匹配长度清零<br> （2）入栈长度不变，出栈长度+2，右括弧不匹配长度清零<br>思路都不能准确判断在读入当前字符时，长度到底应该是多少。<br>3、只考虑到了可以以不匹配的右括号作为分隔，但没想到这个分隔除了长度清零还可以有其他作用<br>4、入栈的只是左括号，根本不知道这个左括号是第几位</p>
<p>正确的思路：<br>1、左括号，则入栈其下标<br>2、右括号，判断<br> （1）如果不匹配，则记录当前位置，结束本次循环<br> （2）如果匹配，则出栈，出栈后判断是否栈空：<br> （2.1）若栈空，说明从该处到上次不匹配的位置的串是合法的括号串，那么当前长度为即为上次不匹配的右括号的后一位到当前 （2.2）若栈不空，说明前面有未匹配的左括号，但是当前的右括号匹配上的是刚刚出栈的左括号，两者之间一定是合法的括号串，那么当前长度就是，出栈的左括号到当前右括号（注意，有一个规律：如果堆栈中有挨着的左括号，那么它们的下标一定是连续的，不过不知道这个规律的话，其实也可以在pop前先得到出栈的下标）<br> 另，我们统计的是包含当前位置的全部合法串长，所以覆盖了全部情况。<br>3、关键点：记录下标</p>
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<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br></pre></td><td class="code"><pre><span class="line">public class Solution {</span><br><span class="line"> private char[] stack = new char[100];</span><br><span class="line"> private int top = -1;</span><br><span class="line"> </span><br><span class="line"> private void push(char x){</span><br><span class="line"> stack[++top] = x;</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> private void pop(){</span><br><span class="line"> top--;</span><br><span class="line"> }</span><br><span class="line"> public boolean isValid(String s) {</span><br><span class="line"> char[] str = s.toCharArray();</span><br><span class="line"> int length = str.length;</span><br><span class="line"> try{</span><br><span class="line"> for(int i = 0; i < length ;i++) {</span><br><span class="line"> switch(str[i]){</span><br><span class="line"> case '(':</span><br><span class="line"> push(str[i]);</span><br><span class="line"> break;</span><br><span class="line"> case '[':</span><br><span class="line"> push(str[i]);</span><br><span class="line"> break;</span><br><span class="line"> case '{':</span><br><span class="line"> push(str[i]);</span><br><span class="line"> break;</span><br><span class="line"> case ')':</span><br><span class="line"> if(stack[top] == '(') {</span><br><span class="line"> pop();</span><br><span class="line"> }</span><br><span class="line"> else {</span><br><span class="line"> return false; </span><br><span class="line"> }</span><br><span class="line"> break;</span><br><span class="line"> case ']':</span><br><span class="line"> if(stack[top] == '[') {</span><br><span class="line"> pop();</span><br><span class="line"> }</span><br><span class="line"> else {</span><br><span class="line"> return false; </span><br><span class="line"> }</span><br><span class="line"> break;</span><br><span class="line"> case '}':</span><br><span class="line"> if(stack[top] == '{') {</span><br><span class="line"> pop();</span><br><span class="line"> }</span><br><span class="line"> else {</span><br><span class="line"> return false; </span><br><span class="line"> }</span><br><span class="line"> break;</span><br><span class="line"> default:</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line"> } </span><br><span class="line"> }</span><br><span class="line"> catch(Exception e) {</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line"> if(top == -1) {</span><br><span class="line"> return true;</span><br><span class="line"> }</span><br><span class="line"> else {</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>简单的栈操作。<br>parenthesis[pə’rɛnθəsɪs]圆括号</p>
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81. Search in Rotated Sorted Array II
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<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line">public class Solution {</span><br><span class="line"> public boolean search(int[] nums, int target) {</span><br><span class="line"> if(nums == null) {</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line"> int length = nums.length;</span><br><span class="line"> for(int i = 0; i < length; i++) {</span><br><span class="line"> if(nums[i] == target) {</span><br><span class="line"> return true; </span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>这个题和33题，都可以改为二分查找，时间复杂度会从O(n)降到O(logn)，之后仔细考虑</p>
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33. Search in Rotated Sorted Array
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<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br></pre></td><td class="code"><pre><span class="line">public class Solution {</span><br><span class="line"> public int search(int[] nums, int target) {</span><br><span class="line"> if(nums == null) {</span><br><span class="line"> return -1;</span><br><span class="line"> }</span><br><span class="line"> int length = nums.length;</span><br><span class="line"> for(int i = 0; i < length; i++) {</span><br><span class="line"> if(nums[i] == target) {</span><br><span class="line"> return i; </span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return -1;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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82. Remove Duplicates from Sorted List II
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<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br></pre></td><td class="code"><pre><span class="line">/**</span><br><span class="line"> * Definition for singly-linked list.</span><br><span class="line"> * public class ListNode {</span><br><span class="line"> * int val;</span><br><span class="line"> * ListNode next;</span><br><span class="line"> * ListNode(int x) { val = x; }</span><br><span class="line"> * }</span><br><span class="line"> */</span><br><span class="line">public class Solution {</span><br><span class="line"> public ListNode deleteDuplicates(ListNode head) {</span><br><span class="line"> if(head == null ||head.next == null){</span><br><span class="line"> return head;</span><br><span class="line"> }</span><br><span class="line"> ListNode p = head.next;</span><br><span class="line"> ListNode result = new ListNode(-1);</span><br><span class="line"> ListNode previous = result;</span><br><span class="line"> ListNode current = head;</span><br><span class="line"> while(p != null) {</span><br><span class="line"> if(current.val == p.val){</span><br><span class="line"> while(p != null && current.val == p.val) {</span><br><span class="line"> p = p.next;</span><br><span class="line"> }</span><br><span class="line"> previous.next = p;</span><br><span class="line"> current = p;</span><br><span class="line"> if(p != null) {</span><br><span class="line"> p = p.next;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> else if(current.val != p.val) {</span><br><span class="line"> current.next = p;</span><br><span class="line"> previous.next = current;</span><br><span class="line"> current = current.next;</span><br><span class="line"> previous = previous.next;</span><br><span class="line"> p = p.next;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return result.next;</span><br><span class="line"> }</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>因为要把重复的元素都删掉，所以需要两个指针，一个指向当前节点，一个指向当前节点的前一个节点。<br>但由于有可能前几个元素重复，当前节点和当前节点前一个节点的关系不好处理，考虑增加一个额外的头节点，作为初始状态时当前节点的前一个节点。</p>
<p>最开始没想到加头结点，想通过标记重复时的元素值，来判断当前节点是否重复，逻辑有点混乱，有点把自己绕晕。</p>
<p>之后想到加头结点，并且不考虑当前指针回退的问题，在循环中再套一个循环，只要当前节点和p指向的节点（下1,2,3，……n个节点）重复，就让p继续向后移直到为空格或找到不重复的节点，这样逻辑就清晰了许多。</p>
<p>最开始的混乱主要在于创建结果链表时，增加的节点之后是有可能被删除的，就会出现当前指针回退的问题，就容易逻辑混乱。</p>
<p>另，这个题目有过改动，最初题目限制了最多元素重复两次，现在的题目无此限制，应理解为重复任意次数。</p>
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9. Palindrome Number
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<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br></pre></td><td class="code"><pre><span class="line">public class Solution {</span><br><span class="line"> </span><br><span class="line"> public boolean isPalindrome(int x) {</span><br><span class="line"> int length = 1;</span><br><span class="line"> long tmp = 9;</span><br><span class="line"> if( x == -2147483648) {</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line"> if(x < 0) {</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line"> while(x > tmp) {</span><br><span class="line"> length++;</span><br><span class="line"> tmp = tmp * 10 + 9;</span><br><span class="line"> }</span><br><span class="line"> return judge(x,length);</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> public boolean judge(int x,int length) {</span><br><span class="line"> int first = x / pow(10,length - 1); </span><br><span class="line"> int last = x % 10;</span><br><span class="line"> if(first != last) {</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line"> else if(first == last && (length == 2 || length == 1)) {</span><br><span class="line"> return true;</span><br><span class="line"> }</span><br><span class="line"> else {</span><br><span class="line"> int next_x = (x - first * pow(10,length - 1) - last) / 10;</span><br><span class="line"> return judge(next_x,length - 2);</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> </span><br><span class="line"> public int pow(int base, int length){</span><br><span class="line"> int result = 1;</span><br><span class="line"> if(length == 0) {</span><br><span class="line"> return result;</span><br><span class="line"> }</span><br><span class="line"> for(int i = 0; i < length; i++) {</span><br><span class="line"> result = result * base;</span><br><span class="line"> }</span><br><span class="line"> return result;</span><br><span class="line"> } </span><br><span class="line"> </span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>判断回文数字，考虑使用递归<br>1、每次判断传入数字的首位和末位是否相等<br>2、若不等返回false<br>3、若相等且数字长度为1或2，则返回ture<br>3、若相等且数字长度大于2，将本数字首位和末位去掉，递归调用</p>
<p>这里认为负数都不是回文需要注意，题目只是提示这个问题要注意，但并没有说怎么处理，应该属于一种默认。</p>
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