01 2021 档案
摘要:连接所有点的最小费用 class Solution { public int minCostConnectPoints(int[][] points) { //prim算法,时间复杂度 n^2 int res = 0, n = points.length; int[][] g = new int[n
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摘要:int l = 0, r = nums.length - 1; while (l <= r) { int mid = (r - l) / 2 + l; if (nums[l] <= nums[mid]) { //ans = mid; r = mid - 1; } else { l = mid + 1
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摘要:最小覆盖子串 class Solution { public String minWindow(String s, String t) { int[] a = new int[128]; for (char ch : t.toCharArray()) { a[ch]++; } int[] b = n
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摘要:水域大小 这里图也是有含义的,每次搜索的是该点的周围8个。(节点有i * j个) class Solution { private int[][] dirs = new int[][]{{1, 1}, {1, -1}, {1, 0}, {-1, 0}, {-1, -1}, {-1, 1}, {0,
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