/** * @param {number[]} nums * @return {string}*/varlargestNumber =function(nums) { nums.sort(function(a,b){returnparseInt(b+""+a) - parseInt(a+""+b); })varrt = nums.join("");returnrt.charAt(0)=="0"?"0":rt; }; 说明 JavaScript中提供了数组的排序方法 arrayObject.sort(so...
然后考虑排序后数组首字母为0的情况。逐个对排序后数组元素相加即可。 classSolution {public:staticboolcmp(inta,intb) {strings1 =to_string(a);strings2 =to_string(b);returns1 + s2 > s2 +s1; }stringlargestNumber(vector<int>&nums) {stringres;if(nums.empty())returnres; sort(nums.begin(), n...
For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330. Note: The result may be very large, so you need to return a string instead of an integer. Credits: Special thanks to @ts for adding this problem and creating all test cases. 这道题和剑指offer读书笔记:第...
给定整数数组nums和整数k,请返回数组中第k个最大的元素。 请注意,你需要找的是数组排序后的第k个最大的元素,而不是第k个不同的元素。 你必须设计并实现时间复杂度为O(n)的算法解决此问题。 示例1: 输入:[3,2,1,5,6,4],k = 2输出:5 示例2: 输入:[3,2,3,1,2,4,5,5,6],k = 4输出:4 ...
In a given integer array nums, there is always exactly one largest element. Find whether the largest element in the array is at least twice as much as every other number in the array. If it is, return the index of the largest element, otherwise return -1. ...
first, second = nums[0], float('inf') for i in range(1, n): num = nums[i] if num > second: return True if num > first: second = num else: first = num return False 55. 跳跃游戏:给定一个非负整数数组 nums ,你最初位于数组的第一个下标。数组中的每个元素代表你在该位置可以跳跃...
【题目】Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal. Note: Each of the array element will not exceed 100. The array size will not exceed 200. Example 1: ...
The array is only modifiable by the update function. You may assume the number of calls to update and sumRange function is distributed evenly. 【解答】写的代码看起来有点啰嗦,大致思路是线段树。也就是说,每个节点都存放一个 sum,这样在求一段区间的和的时候会比较快;而在更新某一个数...
https://leetcode.cn/contest/weekly-contest-335/problems/kth-largest-sum-in-a-binary-tree/ 题意 层次遍历 思路 直接利用BFS将二叉树中所有层次的和求出来,然后将其排序取最大的第k的数即可。 复杂度分析: 时间复杂度:O(n),其中n为二叉树节点的数目。
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