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class Solution { //一个映射表，第二个位置是"abc“,第三个位置是"def"。。。 //这里也可以用map，用数组可以更节省点内存 String[] letter_map = {" ","*","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"}; public List<String> letterCombinations(String digits) { //注意边界条件 if(digits==null |.

class Solution { List<List<Integer>> res; List<Integer> path; public List<List<Integer>> permute(int[] nums){ res = new ArrayList<>(); path = new ArrayList<>(); int[] visited = .

class Solution { public int missingNumber(int[] nums) { int i = 0, j = nums.length - 1; while(i <= j) { int m = (i + j) / 2; if(nums[m] == m) i = m + 1; else j = m - 1; } return i;.

class Solution { public ListNode removeNthFromEnd(ListNode head, int n) { ListNode dummy = new ListNode(0); dummy.next = head; ListNode node1 = dummy; ListNode node2 = dummy; while (node1 != null) { .

给定一个字符串数组 words，请计算当两个字符串 words[i] 和 words[j] 不包含相同字符时，它们长度的乘积的最大值。假设字符串中只包含英语的小写字母。如果没有不包含相同字符的一对字符串，返回 0。class Solution { public int maxProduct(String[] words) { int len = words.length; int[] masks = new int[len]; for(int i=0

class LRUCache { class Node{ int k,v; Node l,r; public Node(int _k, int _v) { k = _k; v = _v; } } int size; Map<Integer, Node> map; Node head, tail; public LRUCache(int .