leetcode-cn.com/problem Givenanintegerarraynums,returnthenumberofreversepairsinthearray. Areversepairisapair(i,j)where0<=i<j<nums.lengthandnums[i]>2*nums[j]. 给定一个数组 nums ,如果 i < j 且 nums[i] > 2*nums[j] 我们就将 (i, j) 称作一个重要翻转对。 你需要返回给定数组中的重要...
1classSolution {2public:3voidreverseString(vector<char>&s) {4if(s.size()==0)return;5intlength = s.size();//获取数组长度6char*front = &s[0];//获取数组第一个元素的地址7char*tail = &s[0];8for(inti =0,j=length-1; i <=length/2,j>= length /2;i++,j--)9{10swap(front[...
说明已经越界了整体使用了尾递归清晰明了Scala 代码object Solution { def reverse(x: Int): Int =...
给定一个整数数组nums,将数组中的元素向右轮转k个位置,其中k是非负数。 示例1: 输入:nums = [1,2,3,4,5,6,7], k = 3输出:[5,6,7,1,2,3,4]解释:向右轮转 1 步:[7,1,2,3,4,5,6]向右轮转 2 步:[6,7,1,2,3,4,5]向右轮转 3 步:[5,6,7,1,2,3,4] ...
Related problem: Reverse Words in a String II 题目标签:Array 题目给了我们一个数组 和 k。 让我们 旋转数组 k 次。 这里有一个很巧妙的方法: 利用数组的length - k 把数组 分为两半; reverse 左边和右边的数组; reverse 总数组。 举一个例子: ...
publicvoidrotate(int[]nums,int k){if(nums==null||nums.length<2){return;}k=k%nums.length;reverse(nums,0,nums.length-k-1);reverse(nums,nums.length-k,nums.length-1);reverse(nums,0,nums.length-1);}privatevoidreverse(int[]nums,int i,int j){int tmp=0;while(i<j){tmp=nums[i];num...
For example, given array S = {-1 2 1 -4}, and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2). 【解答】先排序,然后左右各一指针,在 sum 小于目标值的时候左指针向右走,在 sum 大于目标值的时候右指针向左走。 代码语言:javascript 代码运行次数:0 运行 ...
You are given a binary tree in which each node contains an integer value.Find the number of paths that sum to a given value.The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes). The tree has no...
int N; long[] tr; public int reversePairs(int[] nums) { List<Long> ys = new ArrayList(); for(int i: nums) {//离散化 ys.add((long)i); ys.add((long)i * 2); } Collections.sort(ys);//排序 ys = unique(ys);//去重 N = ys.size(); tr = new long[N + 1];//树状数组...
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