leetcode1289. Minimum Falling Path Sum II

Posted 2021-03-23 seyjs

题目如下：

Given a square grid of integers arr, a falling path with non-zero shifts is a choice of exactly one element from each row of arr, such that no two elements chosen in adjacent rows are in the same column.

Return the minimum sum of a falling path with non-zero shifts.

Example 1:

Input: arr = [[1,2,3],[4,5,6],[7,8,9]]
Output: 13
Explanation: 
The possible falling paths are:
[1,5,9], [1,5,7], [1,6,7], [1,6,8],
[2,4,8], [2,4,9], [2,6,7], [2,6,8],
[3,4,8], [3,4,9], [3,5,7], [3,5,9]
The falling path with the smallest sum is [1,5,7], so the answer is 13.

Constraints:

• 1 <= arr.length == arr[i].length <= 200
• -99 <= arr[i][j] <= 99

解题思路：记dp[i][j]为第i行取第j个元素时，在0~i行区间内获得的最小值。那么显然有dp[i][j] = min(dp[i][j],dp[i-1][k] + arr[i][j]) ,其中j != k。这样的话时间复杂度是O(n)，超时了。再仔细想想，其实对于任意一个j，在dp[i-1]中只要找出最小值即可，当然最小值的所在的列不能和j相同。那么只需要记录dp[i-1]行中的最小值和次小值，如果最小值的下标和j相同就取次小值，否则取最小值。

代码如下：

class Solution(object):
    def minFallingPathSum(self, arr):
        """
        :type arr: List[List[int]]
        :rtype: int
        """
        dp = [[float(‘inf‘)] * len(arr) for _ in arr]
        for i in range(len(arr)):
            dp[0][i] = arr[0][i]

        def getMin(arr):
            min_val = float(‘inf‘)
            min_inx = 0
            for i in range(len(arr)):
                if min_val > arr[i]:
                    min_val = arr[i]
                    min_inx = i
            return (min_val,min_inx)

        def getSecMin(arr,min_inx):
            sec_min_val = float(‘inf‘)
            sec_min_inx = 0
            for i in range(len(arr)):
                if i == min_inx:continue
                if sec_min_val > arr[i]:
                    sec_min_val = arr[i]
                    sec_min_inx = i
            return (sec_min_val,sec_min_inx)

        for i in range(1,len(arr)):
            min_val, min_inx = getMin(dp[i-1])
            sec_min_val, sec_min_inx = getSecMin(dp[i-1],min_inx)
            for j in range(len(arr)):
                if j == min_inx:
                    dp[i][j] = min(dp[i][j],dp[i-1][sec_min_inx] + arr[i][j])
                else:
                    dp[i][j] = min(dp[i][j], dp[i - 1][min_inx] + arr[i][j])
        return min(dp[-1])