<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Mr_hg的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/Mr_hg</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Mr_hg]]></copyright><item><title><![CDATA[leetcode11. Container With Most Water]]></title><link>https://blog.youkuaiyun.com/Mr_hg/article/details/54946579</link><guid>https://blog.youkuaiyun.com/Mr_hg/article/details/54946579</guid><author>Mr_hg</author><pubDate>Thu, 09 Feb 2017 14:04:49 +0800</pubDate><description><![CDATA[public int maxArea(int[] height) {
        int i=0,j=height.length-1;
        int max=0;
        while(i<j){
            max=Math.max(max, Math.min(height[i], height[j])*(j-i));
            if(height[i]]></description><category></category></item><item><title><![CDATA[leetcode6. ZigZag Conversion]]></title><link>https://blog.youkuaiyun.com/Mr_hg/article/details/54946101</link><guid>https://blog.youkuaiyun.com/Mr_hg/article/details/54946101</guid><author>Mr_hg</author><pubDate>Thu, 09 Feb 2017 12:53:13 +0800</pubDate><description><![CDATA[public String convert(String s, int numRows) {
        String[] str=new String[numRows];
        Arrays.fill(str, "");
        int i=0,j;
        while(i<s.length()){
            for(j=0;j<numRows&&i<s]]></description><category></category></item><item><title><![CDATA[leetcode5. Longest Palindromic Substring]]></title><link>https://blog.youkuaiyun.com/Mr_hg/article/details/54945937</link><guid>https://blog.youkuaiyun.com/Mr_hg/article/details/54945937</guid><author>Mr_hg</author><pubDate>Thu, 09 Feb 2017 12:22:17 +0800</pubDate><description><![CDATA[题意找出一个字符串的最长回文子字符串分析字符串最长为1000,比较短，我们可以采用动态规划的思想。 
dp[i][j]表示从下标为i到下标为j的字符串是否为回文串 
dp[i][j]=true if s[i]=s[j]且dp[i+1][j-1]=true    public String longestPalindrome(String s) {
        int n=s.length();]]></description><category></category></item><item><title><![CDATA[leetcode 2.Add Two Numbers]]></title><link>https://blog.youkuaiyun.com/Mr_hg/article/details/54928757</link><guid>https://blog.youkuaiyun.com/Mr_hg/article/details/54928757</guid><author>Mr_hg</author><pubDate>Wed, 08 Feb 2017 16:40:02 +0800</pubDate><description><![CDATA[题意：两个链表相加，每个链表代表一个逆序排列的数字。如2->3->5加3->5表示532+53得5->8->5
  
  分析:同时遍历两个链表，相加，注意进位。
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode re=new ListNode(0);
        ListNode cur]]></description><category></category></item><item><title><![CDATA[leetcode 1. Two Sum]]></title><link>https://blog.youkuaiyun.com/Mr_hg/article/details/54928210</link><guid>https://blog.youkuaiyun.com/Mr_hg/article/details/54928210</guid><author>Mr_hg</author><pubDate>Wed, 08 Feb 2017 16:12:31 +0800</pubDate><description><![CDATA[题目给定一个数组，和一个目标值，返回数组中两个和为目标值的下标，假设结果只有一个分析使用hashMap存储数组元素值和下标，当map中含有target-num[i]的键时，说明已查找到。    public int[] twoSum(int[] nums, int target) {
        int[] re=new int[2];
        HashMap<Integer, Inte]]></description><category></category></item><item><title><![CDATA[390. Elimination Game]]></title><link>https://blog.youkuaiyun.com/Mr_hg/article/details/54633460</link><guid>https://blog.youkuaiyun.com/Mr_hg/article/details/54633460</guid><author>Mr_hg</author><pubDate>Sat, 21 Jan 2017 11:24:39 +0800</pubDate><description><![CDATA[public int lastRemaining(int n) {
        int remain=n;
        int head=1;
        int step=1;
        boolean left=true;
        while(remain>1){
            if(left||(remain%2)==1){]]></description><category></category></item><item><title><![CDATA[leetcode 152 Maximum Product Subarray]]></title><link>https://blog.youkuaiyun.com/Mr_hg/article/details/54599549</link><guid>https://blog.youkuaiyun.com/Mr_hg/article/details/54599549</guid><author>Mr_hg</author><pubDate>Wed, 18 Jan 2017 11:15:21 +0800</pubDate><description><![CDATA[leetcode 152 Maximum Product Subarray
  题目大意：数组中找到一个连续子串，使其乘积最大。数组中至少含有一个元素。
  
  思路：采用动态规划的思想。因为数组中含有负数，且多个负数相乘可能出现为正的情况，所以在动态规划中我们需要记录最小负数。pamin[i]代表子串最后一个元素为nums[i]的最小乘积，pamax[i]代表子串中最后一个元素为nums[i]的]]></description><category></category></item><item><title><![CDATA[leetcode 179 largest number]]></title><link>https://blog.youkuaiyun.com/Mr_hg/article/details/54599226</link><guid>https://blog.youkuaiyun.com/Mr_hg/article/details/54599226</guid><author>Mr_hg</author><pubDate>Wed, 18 Jan 2017 10:43:20 +0800</pubDate><description><![CDATA[leetcode 179 largest number题目
  一个非负数组，通过重新组合，使其能得到最大数。如：[3, 30, 34, 5, 9],最大数是9534330.
分析
  比较两个相邻数字谁在前谁在后，自定义排序方法。我们可以将其转为字符串进行连接进行比较。如数字a和b，可以转换为ab和ba，比较大小。 
  再使用简单的冒泡排序，进行递减排序。 
  注意特殊情况，输入全为0，结果应]]></description><category></category></item></channel></rss>