[leetcode] 62 Unique Paths (Medium)
原题链接
字母题 : unique paths Ⅱ
思路:
dp[i][j]保存走到第i,j格共有几种走法。
因为只能走→或者↓,所以边界条件dp[0][j]+=dp[0][j-1]
同时容易得出递推 dp[i][j]+=dp[i-1][j]+dp[i][j-1]
class Solution {
public:
int uniquePaths(int m, int n) {
if (m == 0 || n == 0) {
return 0;
}
vector<vector<int>> dp(m, vector<int>(n, 0));
dp[0][0] = 1;
for (int j = 1; j < n; j++) dp[0][j] += dp[0][j - 1];
for (int i = 1; i < m; i++) {
for (int j = 0; j < n; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}
};
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