代码随想录day38

62.

模拟一遍就能推出规律来，注意数组越界问题。

/*
 * @lc app=leetcode.cn id=62 lang=cpp
 *
 * [62] 不同路径
 */

// @lc code=start
#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int>>dp(m,vector<int>(n,0));
        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];
    }
};
// @lc code=end

 63.

障碍相当于循环的终止条件。

/*
 * @lc app=leetcode.cn id=63 lang=cpp
 *
 * [63] 不同路径 II
 */

// @lc code=start
#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m=obstacleGrid.size();
        int n=obstacleGrid[0].size();
        if(obstacleGrid[0][0]==1||obstacleGrid[m-1][n-1]==1){
            return 0;
        }
        vector<vector<int>>dp(m,vector<int>(n,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++) {
                if (obstacleGrid[i][j] == 1) continue;
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m-1][n-1];
    }
};
// @lc code=end