Given a square array of integers A
, we want the minimum sum of a falling path through A
.
A falling path starts at any element in the first row, and chooses one element from each row. The next row‘s choice must be in a column that is different from the previous row‘s column by at most one.
Example 1:
Input: [[1,2,3],[4,5,6],[7,8,9]] Output: 12 Explanation: The possible falling paths are:
[1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9]
[2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9]
[3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]
The falling path with the smallest sum is [1,4,7]
, so the answer is 12
.
Note:
1 <= A.length == A[0].length <= 100
-100 <= A[i][j] <= 100
大概都知道求矩阵的最小和,这里不过是规则变了(需要是向下且不同行)
那个dp[i][j]可能从正上方,或者是右上,左上 + A[i][j]得来,考虑边界
class Solution { public: int minFallingPathSum(vector<vector<int>>& A) { if(A.size() == 0) return 0; int inf = 100000; int n = A.size(); vector<vector<int>> dp(n + 10, vector<int>(n + 10, inf)); for (int i = 0; i <= n; ++i) dp[0][i] = A[0][i]; for (int i = 1; i < n; ++i) { for (int j = 0; j < n; ++j) { if (j > 0) dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + A[i][j]); if (j + 1 < n) dp[i][j] = min(dp[i][j], dp[i - 1][j + 1] + A[i][j]); dp[i][j] = min(dp[i][j], dp[i - 1][j] + A[i][j]); } } int ret = inf; for (int x : dp[n - 1]) ret = min(ret, x); return ret; } };
原文地址:https://www.cnblogs.com/yinghualuowu/p/9865121.html
时间: 2024-10-09 14:12:11