62.不同路径
五部曲：
- dp数组下标及含义：dp[i][j]表示从(0,0)到(i,j)位置有几条路径
- dp数组初始化：dp[i][0]=1，dp[0][j]=1
- 递推公式：dp[i][j] = dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
- 遍历方向：从左到右一层一层遍历
- dp数组推到举例：
1 1 1 1 1 1 1
1 2 3 4 5 6 7
1 3 6 10 15 21 28
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];
}
};
63. 不同路径 II
五部曲：
- dp数组下标及含义：dp[i][j]表示从(0,0)到(i,j)位置有几条路径
- dp数组初始化：dp[i][0]=1，dp[0][j]=1,如果有障碍，那么本行或列障碍之后的部分都为0
- 递推公式：dp[i][j] = dp[i][j] = dp[i - 1][j] + dp[i][j - 1]，障碍位置为0
- 遍历方向：从左到右一层一层遍历
- dp数组推到举例：
1 1 1
1 0 1
1 1 2
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 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];
}
};