1 class Solution: 2 def minCostToMoveChips(self, chips: List[int]) -> int: 3 n = len(chips) 4 odd,even = 0,0 5 for i in range(n): 6 cur = chips[i] 7 if cur % 2 == 0: 8 even += 1 9 else: 10 odd += 1 11 return min(even,odd)
1 class Solution: 2 def minCostToMoveChips(self, chips: List[int]) -> int: 3 n = len(chips) 4 odd,even = 0,0 5 for i in range(n): 6 cur = chips[i] 7 if cur % 2 == 0: 8 even += 1 9 else: 10 odd += 1 11 return min(even,odd)
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【LeetCode】39. 组合总和
【LeetCode】705. 设计哈希集合
【LeetCode】706. 设计哈希映射
置上所需要的 最小代价 。示例 1:输入:position = [1,2,3]输出:1
题目:原题链接(简单)解法时间复杂度空间复杂度执行用时
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