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1 public class Solution { 2 public int findMaxForm(String[] strs, int m, int n) { 3 if (strs == null || strs.length == 0) { 4 return 0; 5 } 6 int[][] dp ... 阅读全文
posted @ 2019-06-11 20:30
Sempron2800+
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1 class Solution: 2 def wiggleMaxLength(self, nums: List[int]) -> int: 3 n = len(nums) 4 if n == 0: 5 return 0 6 up,down = 1,1 7 for i in range(... 阅读全文
posted @ 2019-06-11 16:07
Sempron2800+
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第一种思路,时间复杂度O(n^2): 第二种思路,时间复杂度O(n): 阅读全文
posted @ 2019-06-11 15:53
Sempron2800+
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对比leetcode198: 阅读全文
posted @ 2019-06-11 07:03
Sempron2800+
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