Leetcode 1769. Minimum Number of Operations to Move All Balls to Each Box [Python]

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从结果来看， boxes中的任意一位index中的元素如果为1，则向右边挪动到index+1的cost是1，index+2的cost是2，直到Len(boxes) - 1。同理此位置的1向左边挪动也是同理的cost产生。于是从左向右遍历一次，找到一个1，则往结果数组中以找到1的index+1为起始点，顺序增加cost。从左往右的cost计算则把res数组和boxes数组反过来，重复以上过程。

class Solution:def minOperations(self, boxes: str) -> List[int]:N = len(boxes)res = [0]*Nfor index, num in enumerate(boxes):if num == "1" and index != N-1:add = 1for j in range(index+1, N):res[j] += addadd += 1boxes = boxes[::-1]res = res[::-1]for index, num in enumerate(boxes):if num == "1" and index != N-1:add = 1for j in range(index+1, N):res[j] += addadd += 1res = res[::-1]return res

 

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