62/63 Unique Paths

class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> dp(m, vector<int>(n, 1));    //初始化dp数组为全1

for(int i = 1; i < m; ++i) {
for(int j = 1; j < n; ++j) {
dp[j] = dp[i][j-1] + dp[i-1][j];        //DP状态转移方程
}
}
return dp[m-1][n-1];
}
};

1

0

[
[0,0,0],
[0,1,0],
[0,0,0]
]

2

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();            //获取m,n
int n = obstacleGrid[0].size();
int x,y;
vector<vector<int>> dp(m, vector<int>(n, 1));    //初始化
for(int i = 0; i < m; ++i) {
if(obstacleGrid[i][0] == 1) {               //当边缘处出现“1”时，其后面所以格子都不可走到
for(int j = i; j < m; ++j) {
dp[j][0] = 0;
}
break;
}
}
for(int i = 0; i < n; ++i) {                  //同上进行初始化边缘
if(obstacleGrid[0][i] == 1) {
for(int j = i; j < n; ++j) {
dp[0][j] = 0;
}
break;
}
}

for(int i = 1; i < m; ++i) {
for(int j = 1; j < n; ++j) {
x = obstacleGrid[i][j-1] == 1? 0 : 1;    //DP状态转移方程，多加一个判断条件
y = obstacleGrid[i-1][j] == 1? 0 : 1;

dp[i][j] = dp[i][j-1]*x + dp[i-1][j]*y;
}
}
return obstacleGrid[m-1][n-1] == 1? 0 : dp[m-1][n-1];   //最后还需要查看右下角标记是否为“1”
}

};