One element (red) is removed from the input data and inserted in-place into the sorted list with each iteration. Example 1: Input: head = [4,2,1,3] Output: [1,2,3,4] Example 2: Input: head = [-1,5,3,4,0] Output
Given an array withnobjects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively. Note: You are...
// 先排序后调整classSolution{public:voidwiggleSort(vector<int>& nums){if(nums.size() <2)return;sort(nums.begin(),nums.end());// 将数组中一对一对交换for(inti =2; i < nums.size(); i +=2){inttmp = nums[i-1]; nums[i-1] = nums[i]; nums[i] = tmp; } } }; // 贪心...
N is larger than 10000 in only one case. Output For each test case output “Case #x: y1 y2 … yN” (without quotes), where x is the test case number (starting from 1), and yi is the difference of rightmost place and leftmost place of number i. Sample Input 2 3 3 1 2 3 1 ...
(dp[i][2]),方程表示从最早出现的(a[i])开始,后面都只取(a[i]) #pragma GCC optimize(3, "Ofast", "inline") #include <bits/stdc++.h> #define start ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define ll long long #define int ll #define ls st<<1 #define rs st<<...
leetcode笔记:Wiggle Sort 一. 题目描写叙述 Given an unsorted array nums, reorder it in-place such thatnums[0] <= nums[1] >= nums[2] <= nums[3]… For example, givennums = [3, 5, 2, 1, 6, 4], one possible answer is[1, 6, 2, 5, 3, 4]....
count += search(head, nums[i] * 2LL + 1); head = insert(head, nums[i]); }return count; }private: int search(Node* head, long long val) { if (head == NULL) return 0; else if (head->val == val) { return head->count_ge; } else if (head->val > val) { ...