Day1 05-23 动态规划算法练习

​ 这周的重心在算法题上，今天刷了三道算法题，基本都是最优解

1. 不同路径

   题目链接：https://leetcode.cn/problems/unique-paths/

   代码：

class Solution {
    public int[][] dp;
    public int uniquePaths(int m, int n) {
        dp=new int[m][n];
        for (int i = 0; i < m; i++) {
            dp[i][0]=1;
        }
        for (int i = 0; i < n; i++) {
            dp[0][i]=1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
}

2. 不同路径 II

   题目链接：https://leetcode.cn/problems/unique-paths-ii/

   代码：

class Solution {
    public int[][] dp;
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int height=obstacleGrid.length;
        int length=obstacleGrid[0].length;
        dp=new int[height][length];
        if(obstacleGrid[0][0]==0){
            dp[0][0]=1;
        }
        for (int i = 1; i < height; i++) {
            if (obstacleGrid[i][0]==1){
                dp[i][0]=0;
            }else {
                dp[i][0]=dp[i-1][0];
            }
        }
        for (int i = 1; i < length; i++) {
            if (obstacleGrid[0][i]==1){
                dp[0][i]=0;
            }else {
                dp[0][i]=dp[0][i-1];
            }
        }
        for (int i = 1; i < height; i++) {
            for (int j = 1; j < length; j++) {
                if (obstacleGrid[i][j]==1){
                    dp[i][j]=0;
                }else {
                    dp[i][j]=dp[i-1][j]+dp[i][j-1];
                }
            }
        }
        return dp[height-1][length-1];
    }
}

3. 整数拆分

   题目链接：https://leetcode.cn/problems/integer-break/

   代码：

class Solution {
    public int size;
    public int maxValue;
    public int avg;
    public int integerBreak(int n) {
        size=2;
        maxValue=1;
        avg=1;
        for (int i = 2; i < n; i++) {
            int newAvg=(i+1)/(size+1);
            int bigNumberSize=(i+1)-newAvg*(size+1);
            int smallNumberSize=size+1-bigNumberSize;
            int value1=maxValue*(avg+1)/avg;
            int value2= (int) (Math.pow(newAvg,smallNumberSize)*Math.pow((newAvg+1),bigNumberSize));
            if (value1>value2){
                maxValue=value1;
                avg=(i+1)/size;
            }else {
                maxValue=value2;
                size++;
                avg=newAvg;
            }
        }
        return maxValue;
    }
}