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一、Description
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.
题目大意:从一个二维数组的第一行的任意位置开始往下遍历,每次只能到达当前位置的左下、正下或者右下的位置,直到最后一行,使得经过的位置对应的数字总和最小。
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
.
二、Analyzation
可以从数组第二行开始,从第0列起,每个位置加上左上、正上和右上方三个数字中最小的那一个,如果出了边界,则将那个位置的数字置为MAX,如此向下累加,最后一行中最小的数字即为所求。
三、Accepted code
class Solution {
public int minFallingPathSum(int[][] A) {
if (A == null || A.length == 0) {
return 0;
}
int m = A.length;
int n = A[0].length;
for (int i = 1; i < m; i++) {
for (int j = 0; j < n; j++) {
int a, b, c;
if (j == 0) {
a = Integer.MAX_VALUE;
} else {
a = A[i - 1][j - 1];
}
if (j == n - 1) {
c = Integer.MAX_VALUE;
} else {
c = A[i - 1][j + 1];
}
b = A[i - 1][j];
A[i][j] += Math.min(Math.min(a, b), c);
}
}
int min = Integer.MAX_VALUE;
for (int i = 0; i < n; i++) {
if (min > A[m - 1][i]) {
min = A[m - 1][i];
}
}
return min;
}
}