LeetCode 931. Minimum Falling Path Sum
Woodyiiiiiii opened this issue · comments
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
很简单。二维dp数组如下:
class Solution {
public int minFallingPathSum(int[][] A) {
int row = A.length;
int col = A[0].length;
int[][] dp = new int[row][col];
int res = Integer.MAX_VALUE, i = 0, j = 0;
for (i = 0; i < col; ++i) {
dp[0][i] = A[0][i];
}
for (i = 1; i < row; ++i) {
for (j = 0; j < col; ++j) {
if (j == 0) {
dp[i][j] = A[i][j] + Math.min(dp[i - 1][j], dp[i - 1][j + 1]);
}
else if (j == col - 1) {
dp[i][j] = A[i][j] + Math.min(dp[i - 1][j], dp[i - 1][j - 1]);
}
else {
// Math.min(a, b) in java.lang
dp[i][j] = A[i][j] + Math.min(dp[i - 1][j],
Math.min(dp[i - 1][j - 1], dp[i - 1][j + 1]));
}
}
}
for (i = 0; i < col; ++i) {
res = (dp[row - 1][i] < res) ? dp[row - 1][i] : res;
}
return res;
}
}
当然也可以优化成一维数组(需要声明一个变量保存初值),这里不再赘述。
参考资料:
LeetCode原题