62.不同路径
题目链接：62. 不同路径 - 力扣（LeetCode）思路：dp[i][j]表示到（i，j）有dp[i][j]个路径。dp[i][j] = dp[i-1][j] + dp[i][j-1]
class Solution {
public int uniquePaths(int m, int n) {
int[][] 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];
}
}63. 不同路径 II
题目链接：63. 不同路径 II - 力扣（LeetCode）
思路：递归思路和上面一样，只是如果是1则只能取一边
class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int h = obstacleGrid.length;
int l = obstacleGrid[0].length;
int[][] dp = new int[h][l];
if(obstacleGrid[0][0] == 1 || obstacleGrid[h-1][l-1] == 1) return 0;
for(int i = 0; i < h && obstacleGrid[i][0] != 1; i++) dp[i][0] = 1;
for(int i = 0; i < l&& obstacleGrid[0][i] != 1; i++) dp[0][i] = 1;
for(int i = 1; i < h; i++){
for(int j = 1; j < l; j++){
if(dp[i][j] != 1){
if(obstacleGrid[i-1][j] == 1 && obstacleGrid[i][j-1] != 1){
dp[i][j] = dp[i][j-1];
}else if(obstacleGrid[i-1][j] != 1 && obstacleGrid[i][j-1] == 1){
dp[i][j] = dp[i-1][j];
}else if(obstacleGrid[i-1][j] == 1 && obstacleGrid[i][j-1] == 1){
dp[i][j] = 0;
}else{
dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
}
}
return dp[h-1][l-1];
}
}
![[蓝桥杯 2023 省 A] 颜色平衡树：从零开始理解树上莫队 一颗颜色平衡树引发的惨案](https://img-blog.csdnimg.cn/direct/c020e95741a04679a74dc01ba27cf552.png)
