LeetCode 931. Minimum Falling Path Sum

原题链接在这里：https://leetcode.com/problems/minimum-falling-path-sum/

题目：

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. 1 <= A.length == A[0].length <= 100
2. -100 <= A[i][j] <= 100

题解：

For each cell A[i][j], the minimum falling path sum ending at this cell = A[i][j]+ Min(minimum sum ending on its upper left, minimum sum ending on its upper, minimum sum ending on it upper right).

Could use dp to cash previous value.

Time Complexity: O(m*n). m = A.length. n = A[0].length.

Space: O(m*n).

AC Java:

 class Solution {
     public int minFallingPathSum(int[][] A) {
         if(A == null || A.length == 0 || A[0].length == 0){
             return 0;
         }

         int res = Integer.MAX_VALUE;
         int m = A.length;
         int n = A[0].length;
         int [][] dp = new int[m+1][n];

         for(int i = 1; i<=m; i++){
             for(int j = 0; j<n; j++){
                 int leftUp = j==0 ? dp[i-1][j] : dp[i-1][j-1];
                 int rightUp = j == n-1 ? dp[i-1][j] : dp[i-1][j+1];
                 dp[i][j] = A[i-1][j] + Math.min(leftUp, Math.min(dp[i-1][j], rightUp));
                 if(i == m){
                     res = Math.min(res, dp[i][j]);
                 }
             }
         }

         return res;
     }
 }

Could operate on original A.

Time Complexity: O(m*n).

Space: O(1).

AC Java:

 class Solution {
     public int minFallingPathSum(int[][] A) {
         if(A == null || A.length == 0 || A[0].length == 0){
             return 0;
         }

         int res = Integer.MAX_VALUE;
         int m = A.length;
         int n = A[0].length;

         if(m == 1){
             for(int j = 0; j<n; j++){
                 res = Math.min(res, A[0][j]);
             }

             return res;
         }

         for(int i = 1; i<m; i++){
             for(int j = 0; j<n; j++){
                 int leftUp = j==0 ? A[i-1][j] : A[i-1][j-1];
                 int rightUp = j == n-1 ? A[i-1][j] : A[i-1][j+1];
                 A[i][j] += Math.min(leftUp, Math.min(A[i-1][j], rightUp));
                 if(i == m-1){
                     res = Math.min(res, A[i][j]);
                 }
             }
         }

         return res;
     }
 }