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931. Minimum Falling Path Sum
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:
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1 ⇐ A.length == A[0].length ⇐ 100 -
-100 ⇐ A[i][j] ⇐ 100
思路分析
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/**
* 动态规划，没有优化
*
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-03 14:26:47
*/
public int minFallingPathSum(int[][] matrix) {
int[][] dp = new int[matrix.length][matrix[0].length];
System.arraycopy(matrix[0], 0, dp[0], 0, matrix[0].length);
for (int row = 1; row < matrix.length; row++) {
for (int column = 0; column < matrix[row].length; column++) {
int num = dp[row - 1][column];
if (0 <= column - 1) {
num = Math.min(num, dp[row - 1][column - 1]);
}
if (column + 1 < matrix[row].length) {
num = Math.min(num, dp[row - 1][column + 1]);
}
dp[row][column] = matrix[row][column] + num;
}
}
int result = Integer.MAX_VALUE;
for (int column = 0; column < dp[dp.length - 1].length; column++) {
result = Math.min(result, dp[dp.length - 1][column]);
}
return result;
}