classSolution {public:intfirstMissingPositive(intA[],intn) {if(n ==0)return1;//if A[i] is negative, i+1 exists in original A//partition, non-negative onlyintlow =0;inthigh = n-1;intend = n-1;while(low <=high) {
Given an unsorted integer array, find the first missing positive integer. For example, Given[1,2,0]return3, and[3,4,-1,1]return2. Your algorithm should run inO(n) time and uses constant space. 通过swap操作,把各个元素swap到相应的位置上,然后再扫描一遍数组,确定丢失的正数的位置 1classSolut...
Given an unsorted integer arraynums, return the smallest missing positive integer. You must implement an algorithm that runs inO(n)time and uses constant extra space. Example 1: Input: nums = [1,2,0] Output: 3 Example 2: Input: nums = [3,4,-1,1] Output: 2 Example 3: Input: nums...
为了减少时间复杂度,可以把 input 数组都装到 map 中,然后 i 循环从 1 开始,依次比对 map 中是否存在 i,只要不存在 i 就立即返回结果,即所求。 参考代码 packageleetcodefuncfirstMissingPositive(nums[]int)int{numMap:=make(map[int]int,len(nums))for_,v:=rangenums{numMap[v]=v}forindex:=1;index...
[LeetCode] 41. First Missing Positive ☆☆☆(第一个丢失的正数),Givenanunsortedintegerarray,findthesmallestmissingpositiveinteger.Examp
41. 缺失的第一个正数 First Missing Positive 力扣 LeetCode 题解 56 -- 7:06 App 52. N 皇后 II N-Queens II 力扣 LeetCode 题解 186 -- 3:57 App 49. 字母异位词分组 Group Anagrams 力扣 LeetCode 题解 60 -- 5:43 App 17. 电话号码的字母组合 Letter Combinations of a Phone Number ...
Given an unsorted integer array, find the smallest missing positive integer. 给定一个未排序的整数数组,找出其中缺少的最小的正整数。 Example 1: Input: [1,2,0] Output: 3 Example 2: Input: [3,4,-1,1] Output: 2 Example 3: Input: [7,8,9,11,12] ...
今天是一道在LeetCode上标记为Hard的题目,Acceptance为22.9%的题目,虽然知道思路之后会发现其实较为简单。 题目如下: Given an unsorted integer array, find the first missing positive integer. For example,Given[1,2,0]return3, and[3,4,-1,1]return2. ...
LeetCode:First Missing Positive LeetCode 题目链接:First Missing Positive Given an unsorted integer array, find the first missing positive integer. For example, Given [1,2,0] return 3, and [3,4,-1,1] return 2. Your algorithm should run in O(n) time and uses constant space....
public int firstMissingPositive(int[] nums) { for (int i = 0; i < nums.length; ) { int n = nums[i]; if (n >= 1 && n <= nums.length + 1 && nums[n - 1] != n) {//这个数在答案区间 int tmp = nums[n - 1]; ...