摘要: ①: public void Merge(int[] nums1, int m, int[] nums2, int n) { int size = m + n; for (int i = 0; i < nums2.Length; i++) { if (m < size) { nums1[m++] = 阅读全文
posted @ 2021-11-10 11:49 奕心1999 阅读(18) 评论(0) 推荐(0)
摘要: ①:两次遍历 时间复杂度为O(n平方) public int[] TwoSum(int[] nums, int target) { for (int i = 0; i < nums.Length; i++) { for (int j = i+1; j < nums.Length; j++) { if 阅读全文
posted @ 2021-11-10 11:06 奕心1999 阅读(22) 评论(0) 推荐(0)