<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[X1579983102的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/X1579983102</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; X1579983102]]></copyright><item><title><![CDATA[遍历路径与动态规划]]></title><link>https://blog.youkuaiyun.com/X1579983102/article/details/107155149</link><guid>https://blog.youkuaiyun.com/X1579983102/article/details/107155149</guid><author>X1579983102</author><pubDate>Mon, 06 Jul 2020 12:34:34 +0800</pubDate><description><![CDATA[class Solution {
public:
    int move(vector&lt;vector&lt;int&gt;&gt;&amp; o) {
        int length = o.size(), width = o[0].size();
        if(length==0|| width==0||o[0][0] == 1) return 0;
        vector&lt;vector&lt;int&gt;&gt; temp(length + 1, vector&l..]]></description><category></category></item><item><title><![CDATA[找到乘积最大的子数组]]></title><link>https://blog.youkuaiyun.com/X1579983102/article/details/106931927</link><guid>https://blog.youkuaiyun.com/X1579983102/article/details/106931927</guid><author>X1579983102</author><pubDate>Tue, 23 Jun 2020 21:05:01 +0800</pubDate><description><![CDATA[动态规划:
class Solution {
public:
  int maxProduct(vector&lt;int&gt;&amp; nums) {
  int len = nums.size(), ans = nums[0];
    int prevMin = nums[0], prevMax = nums[0];
    int temp1 = 0, temp2 = 0;
    for (int i = 1; i &lt; len; i++) {
      temp1 = prevMin.]]></description><category></category></item><item><title><![CDATA[前缀和与哈希表]]></title><link>https://blog.youkuaiyun.com/X1579983102/article/details/106916523</link><guid>https://blog.youkuaiyun.com/X1579983102/article/details/106916523</guid><author>X1579983102</author><pubDate>Tue, 23 Jun 2020 20:55:36 +0800</pubDate><description><![CDATA[前缀和解法:
class Solution {
public:
    int subarraySum(vector&lt;int&gt;&amp; nums, int k) {
        int count = 0;
        for (int i = 0; i &lt; nums.size(); i++) { //实现从任意的num[i]加到[j]
            int sum = 0;
            for (int j = i; j&gt;=0; j--) {
  .]]></description><category></category></item><item><title><![CDATA[异或运算符与寻找只出现一次的数字]]></title><link>https://blog.youkuaiyun.com/X1579983102/article/details/106904041</link><guid>https://blog.youkuaiyun.com/X1579983102/article/details/106904041</guid><author>X1579983102</author><pubDate>Tue, 23 Jun 2020 09:14:29 +0800</pubDate><description><![CDATA[异或运算符解法：
class Solution {
public:
    int singleNumber(vector&lt;int&gt;&amp; nums) {
        int res=0;
        for(int i=0;i&lt;nums.size();i++)
        res^=nums[i]; //^:异或运算符号
        return res;
    }
};

原理：
相同的数字两两相消，最后只有单身贵族活着
另解：
class Solution {.]]></description><category></category></item><item><title><![CDATA[二叉树的广度优先搜索]]></title><link>https://blog.youkuaiyun.com/X1579983102/article/details/106902157</link><guid>https://blog.youkuaiyun.com/X1579983102/article/details/106902157</guid><author>X1579983102</author><pubDate>Mon, 22 Jun 2020 15:09:31 +0800</pubDate><description><![CDATA[/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector&lt;vector&lt;int&gt;&gt.]]></description><category></category></item><item><title><![CDATA[栈的操作与栈中检索最小值算法]]></title><link>https://blog.youkuaiyun.com/X1579983102/article/details/106901975</link><guid>https://blog.youkuaiyun.com/X1579983102/article/details/106901975</guid><author>X1579983102</author><pubDate>Mon, 22 Jun 2020 14:59:36 +0800</pubDate><description><![CDATA[重点是getMin()，整块代码为此服务，其他的都能直接调用`
class MinStack {
public:
    stack&lt;int&gt;S;
    stack&lt;int&gt;Min; //记录每次入栈后变化的最小值

    void push(int x) {
        S.push(x);
        if(Min.empty()||x&lt;=Min.top())Min.push(x);
    }
    
    void pop() {
        if.]]></description><category></category></item><item><title><![CDATA[递归二分法算x的n次方]]></title><link>https://blog.youkuaiyun.com/X1579983102/article/details/106054563</link><guid>https://blog.youkuaiyun.com/X1579983102/article/details/106054563</guid><author>X1579983102</author><pubDate>Mon, 11 May 2020 15:03:17 +0800</pubDate><description><![CDATA[递归二分法算x的n次方

class Solution {
public:
    double myPow(double x, long n) {
        if(x==-1){
            if(n%2==0) return 1;
            else return -1;
        }else{
            if(n==0||x==1) return 1;
            if(n==1) return x;
            if(n&l]]></description><category></category></item></channel></rss>