<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[pharos的博客]]></title><description><![CDATA[哈哈哈哈哈哈]]></description><link>https://blog.youkuaiyun.com/ILUU121</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; ILUU121]]></copyright><item><title><![CDATA[Leetcode_279_PerfectSquares_按平方数来分割整数]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101702368</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101702368</guid><author>ILUU121</author><pubDate>Sun, 29 Sep 2019 17:16:41 +0800</pubDate><description><![CDATA[package DP;

public class _279_PerfectSquares {
    /**
     * 1.递归+dfs
     *              13
     *      /1      |4   \9
     *     12       9      4
     *   /1  |4  \9      /1    \4
     *  11   8...]]></description><category></category></item><item><title><![CDATA[Leetcode_343_IntegerBreak_分割整数的最大乘积]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101698432</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101698432</guid><author>ILUU121</author><pubDate>Sun, 29 Sep 2019 16:24:30 +0800</pubDate><description><![CDATA[package DP;

import java.util.List;

public class _343_IntegerBreak {
    // 1. 递归暴力解法
    // F(n) = max{i * F(n-i)}, i = 1,2,3,4,5...n-1
    public int helper1(int n) {
        if (n == 2){
         ...]]></description><category></category></item><item><title><![CDATA[Leetcode_413_ArithmeticSlices_数组中等差递增子区间的个数]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101677831</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101677831</guid><author>ILUU121</author><pubDate>Sun, 29 Sep 2019 10:59:21 +0800</pubDate><description><![CDATA[package DP;

/**
 * 数组中等差递增子区间的个数
 * exm:
 * A = [0, 1, 2, 3, 4]
 *
 * return: 6, for 3 arithmetic slices in A:
 *
 * [0, 1, 2],
 * [1, 2, 3],
 * [0, 1, 2, 3],
 * [0, 1, 2, 3, 4],
 * [ 1, 2, 3, 4],
 *...]]></description><category></category></item><item><title><![CDATA[理解动态规划]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101624917</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101624917</guid><author>ILUU121</author><pubDate>Sat, 28 Sep 2019 17:15:56 +0800</pubDate><description><![CDATA[1. 理解动态规划
1.1 递推！（递归+记忆化）
1.2 状态的定义：opt[n], dp[n], fib[n]
1.3 状态转移方程：opt[n] = best_of(opt[n-1], opt[n-2], …)
1.4 最优子结构
2. 动态规划例子
2.1 斐波那契数列
2.2 count the paths（计算路径数）
2.2.1 递归解法
解释：小人只能向右走，或者向下走。
int ...]]></description><category></category></item><item><title><![CDATA[Leetcode_62_uniquePaths_矩阵的总路径数]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101624590</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101624590</guid><author>ILUU121</author><pubDate>Sat, 28 Sep 2019 17:11:51 +0800</pubDate><description><![CDATA[package DP;

public class _62_uniquePaths {
    //1.  dp
    public int uniquePaths(int m, int n){
        int[][] dp = new int[m][n];
        for (int i = 0; i &lt; m; i++){
            dp[i][0] = 1;...]]></description><category></category></item><item><title><![CDATA[Leetcode_64_minPathSum_矩阵的最小路径和]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101621627</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101621627</guid><author>ILUU121</author><pubDate>Sat, 28 Sep 2019 16:22:17 +0800</pubDate><description><![CDATA[package DP;

public class _64_minPathSum {
    // 1. 递归-dfs
    private int helper(int[][] grid, int i, int j){
        if (i == grid.length - 1 &amp;&amp; j == grid[0].length -1){
            return ...]]></description><category></category></item><item><title><![CDATA[Leetcode_213_HouseRober2_打家劫舍]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101603549</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101603549</guid><author>ILUU121</author><pubDate>Sat, 28 Sep 2019 11:25:58 +0800</pubDate><description><![CDATA[package DP;

import java.util.Arrays;

public class _213_HouseRober2 {
    /**
     * 动态规划，和第198道一样。
     * @param nums
     * @return
     */
    private static int myRob(int[] nums){
        int n =...]]></description><category></category></item><item><title><![CDATA[链表常见面试题]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101601508</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101601508</guid><author>ILUU121</author><pubDate>Sat, 28 Sep 2019 10:48:30 +0800</pubDate><description><![CDATA[import java.util.Stack;

class ListNode{
    int val;
    ListNode next;
    ListNode(int val){
        this.val = val;
    }
}
public class MyLinkListDemo {
    /**
     * 1. 翻转链表
     */
    public ...]]></description><category></category></item><item><title><![CDATA[二叉树常见面试题集合]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101601412</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101601412</guid><author>ILUU121</author><pubDate>Sat, 28 Sep 2019 10:47:02 +0800</pubDate><description><![CDATA[import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

/**
 * 二叉树刷题集锦
 */
class TreeNode{
    int val;
    TreeNode left;
    TreeNode right;
    public TreeNode(){}
    publi...]]></description><category></category></item><item><title><![CDATA[Leetcode_70_climbingStairs_爬楼梯]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101547014</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101547014</guid><author>ILUU121</author><pubDate>Fri, 27 Sep 2019 17:24:52 +0800</pubDate><description><![CDATA[题目描述：
有 N 阶楼梯，每次可以上一阶或者两阶，求有多少种上楼梯的方法。
解题与代码：
package DP;

/**
 * 爬楼梯
 */
public class _70_climbingStairs {
    public static int climbStairs(int n){
        int[] dp = new int[n+1];
        dp[0] = 1...]]></description><category></category></item><item><title><![CDATA[Leetcode__198_rob_强盗抢劫]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/101543617</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/101543617</guid><author>ILUU121</author><pubDate>Fri, 27 Sep 2019 17:15:16 +0800</pubDate><description><![CDATA[题目描述：
抢劫一排住户，但是不能抢邻近的住户，求最大抢劫量。
解决思路和代码：
package DP;

import java.util.Arrays;

/**
 * 强盗抢劫
 * Find recursive relation
 * Recursive (top-down)
 * Recursive + memo (top-down)
 * Iterative + memo (botto...]]></description><category></category></item><item><title><![CDATA[46-permutation]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/90759266</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/90759266</guid><author>ILUU121</author><pubDate>Mon, 03 Jun 2019 21:18:59 +0800</pubDate><description><![CDATA[1. 思路：使用回溯算法
2. 转换成树形问题
3. 


4. 代码
class Solution:
    # 回溯算法
    res = list()  # 保存结果
    used = []
    def generatePermutation(self, nums: list, index: int, p: list):
        if index == len(nums):...]]></description><category></category></item><item><title><![CDATA[131分割回文串]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/90705456</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/90705456</guid><author>ILUU121</author><pubDate>Thu, 30 May 2019 20:48:39 +0800</pubDate><description><![CDATA[class Solution:
    # 回溯

    res = []
    tmp = []
    def isPalindrome(self, s: str):  # 判断是否是回文串
        return s == s[::-1]

    def dfs(self, s: str):
        '''

        :param s:
        :para...]]></description><category></category></item><item><title><![CDATA[动态规划实践]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/90485720</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/90485720</guid><author>ILUU121</author><pubDate>Thu, 23 May 2019 18:29:15 +0800</pubDate><description><![CDATA[leetcode-70 : climbing stairs

class Solution3:
    
    '''
    分析：一次走一阶，也可以一次走两阶。
    到第n阶的总总走法数 = 到第n-1阶的走法数 + 到第n - 2 阶的总走法数。
    '''
    def climbStairs(self, n):
        if n == 0 or n == 1 o...]]></description><category></category></item><item><title><![CDATA[动态规划]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/90449701</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/90449701</guid><author>ILUU121</author><pubDate>Wed, 22 May 2019 18:36:16 +0800</pubDate><description><![CDATA[动态规划理解

]]></description><category></category></item><item><title><![CDATA[关于递归]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/90414358</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/90414358</guid><author>ILUU121</author><pubDate>Tue, 21 May 2019 20:16:19 +0800</pubDate><description><![CDATA[转载自：递归


]]></description><category></category></item><item><title><![CDATA[402-removeKdigits]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/90409821</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/90409821</guid><author>ILUU121</author><pubDate>Tue, 21 May 2019 16:37:47 +0800</pubDate><description><![CDATA['''
使用一个栈作为辅助，遍历数字字符串，当当前的字符比栈最后的字符小的时候，说明要把栈的最后的这个字符删除掉。为什么呢？你想，把栈最后的字符删除掉，然后用现在的字符进行替换，是不是数字比以前的那种情况更小了？所以同样的道理，做一个while循环！这个很重要，可是我没有想到。在每一个数字处理的时候，都要做一个循环，使得栈里面最后的数字比当前数字大的都弹出去。

最后，如果K还没用完，那要删除哪里...]]></description><category></category></item><item><title><![CDATA[169-求众数]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/89972064</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/89972064</guid><author>ILUU121</author><pubDate>Wed, 08 May 2019 20:28:09 +0800</pubDate><description><![CDATA[class Solution1:
    #1.暴力求解
    def majorityElement(self, nums) -&gt; int:
        '''

        :param nums: List[int]
        :return: int
        '''
        Max = 0
        res = nums[0]
        f...]]></description><category></category></item><item><title><![CDATA[Bias PUF实验设计和原理介绍]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/89921533</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/89921533</guid><author>ILUU121</author><pubDate>Tue, 07 May 2019 15:12:32 +0800</pubDate><description><![CDATA[]]></description><category></category></item><item><title><![CDATA[栈的应用之实现简单的计算器]]></title><link>https://blog.youkuaiyun.com/ILUU121/article/details/89501224</link><guid>https://blog.youkuaiyun.com/ILUU121/article/details/89501224</guid><author>ILUU121</author><pubDate>Wed, 24 Apr 2019 19:31:36 +0800</pubDate><description><![CDATA[栈的应用之实现简单的计算器
步骤1：将前缀表达式转换为后缀表达式
步骤2：计算后缀表达式
//
// Created by lenovo on 2019/4/24.
//
//栈的应用，简单计算器
/**
 * 步骤1：中缀表达式转换为后缀表达式
 * 步骤2：计算后缀表达式
 */
#include &lt;iostream&gt;
#include &lt;stack&gt;
#include...]]></description><category></category></item></channel></rss>