代码随想录-035期-算法训练营【博客笔记汇总表】-CSDN博客

第九章 动态规划part02● 62.不同路径 
● 63. 不同路径 II 今天开始逐渐有 dp的感觉了，题目不多，就两个 不同路径，可以好好研究一下详细布置 62.不同路径 本题大家掌握动态规划的方法就可以。 数论方法 有点非主流，很难想到。 https://programmercarl.com/0062.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84.html  
视频讲解：https://www.bilibili.com/video/BV1ve4y1x7Eu63. 不同路径 II https://programmercarl.com/0063.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84II.htmlhttps://programmercarl.com/0063.%E4%B8%8D%E5%90%8C%E8%B7%AF%E5%BE%84II.html  
视频讲解：https://www.bilibili.com/video/BV1Ld4y1k7c6 

目录

0062_不同路径

0063_不同路径II

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0062_不同路径

package com.question.solve.leetcode.programmerCarl2._10_dynamicProgramming;import java.util.Arrays;public class _0062_不同路径 {
}class Solution0062 {//超时public int res = 0;public int uniquePaths(int m, int n) {dfs(m, n, 1, 1);return res;}public void dfs(int m, int n, int i, int j) {if (i > m || j > n) {return;}if (i == m && j == n) {res++;}dfs(m, n, i + 1, j);dfs(m, n, i, j + 1);}
}class Solution0062_2 {//超时public int uniquePaths(int m, int n) {return dfs(1, 1, m, n);}int dfs(int i, int j, int m, int n) {if (i > m || j > n)   //越界了return 0;if (i == m && j == n) //找到一种方法，相当于找到了叶子节点return 1;return dfs(i + 1, j, m, n) + dfs(i, j + 1, m, n);}
}class Solution0062_3 {/*** 1. 确定dp数组下标含义 dp[i][j] 到每一个坐标可能的路径种类* 2. 递推公式 dp[i][j] = dp[i-1][j] dp[i][j-1]* 3. 初始化  dp[i][0]=1、dp[0][i]=1，初始化横竖就可* 4. 遍历顺序，一行一行遍历* 5. 推导结果 。。。。。。。。** @param m* @param n* @return*/public static int uniquePaths(int m, int n) {int[][] dp = new int[m][n];//初始化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];}public int uniquePaths2(int m, int n) {//状态压缩//在二维dp数组中，当前值的计算只依赖正上方和正左方，因此可以压缩成一维数组。int[] dp = new int[n];//初始化，第一行只能从正左方跳过来，所以只有一条路径。Arrays.fill(dp, 1);for (int i = 1; i < m; i++) {//第一列也只有一条路，不用迭代，所以从第二列开始for (int j = 1; j < n; j++) {dp[j] += dp[j - 1];//dp[j] = dp[j] (正上方)+ dp[j - 1] (正左方)}}return dp[n - 1];}public long uniquePaths3(int m, int n) {//数论long numerator = 1; // 分子int denominator = m - 1; // 分母int count = m - 1;int t = m + n - 2;while (count > 0) {numerator *= (t--);while (denominator != 0 && numerator % denominator == 0) {numerator /= denominator;denominator--;}count--;}return numerator;}
}

0063_不同路径II

package com.question.solve.leetcode.programmerCarl2._10_dynamicProgramming;public class _0063_不同路径II {
}class Solution0063 {//超时public int res = 0;public int uniquePathsWithObstacles(int[][] obstacleGrid) {dfs(obstacleGrid, 0, 0);return res;}private void dfs(int[][] obstacleGrid, int i, int j) {if (i >= obstacleGrid.length || j >= obstacleGrid[0].length || obstacleGrid[i][j] == 1) {return;}if (i == obstacleGrid.length - 1 && j == obstacleGrid[0].length - 1) {res++;}dfs(obstacleGrid, i + 1, j);dfs(obstacleGrid, i, j + 1);}
}class Solution0063_2 {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;int[][] dp = new int[m][n];//如果在起点或终点出现了障碍，直接返回0if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {return 0;}for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) {dp[i][0] = 1;}for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {dp[0][j] = 1;}for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {dp[i][j] = (obstacleGrid[i][j] == 0) ? dp[i - 1][j] + dp[i][j - 1] : 0;}}return dp[m - 1][n - 1];}//空间优化版本public int uniquePathsWithObstacles2(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;int[] dp = new int[n];for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {dp[j] = 1;}for (int i = 1; i < m; i++) {for (int j = 0; j < n; j++) {if (obstacleGrid[i][j] == 1) {dp[j] = 0;} else if (j != 0) {dp[j] += dp[j - 1];}}}return dp[n - 1];}
}