LeetCode 41. First Missing Positive(缺失的第一个正数) 题目 题目补充说明:这题中输入的数据中可以有负值,可以有重复的值,其中真正需要的排序效果是1,2,3,4…n这样的效果,让在这些数据之间找到缺失的第一个正数。 解析 分析:在明确了题目的实际要求之后,就可以设计算法了,比如如何处理负数和重复的数据,如何...
swap要一直循环。 classSolution {public:intfirstMissingPositive(vector<int>&nums) {for(inti =0;i < nums.size();i++){if(nums[i] <= nums.size() && nums[i] >0&& nums[nums[i] -1] !=nums[i]) swap(nums[i],nums[nums[i]-1]);}for(inti =0;i < nums.size();i++){if(nums[...
Smallest Missing Non-negative Integer After Operations Maximum Number of Integers to Choose From a Range II 参考资料: https://leetcode.com/problems/first-missing-positive/ https://leetcode.com/problems/first-missing-positive/discuss/17071/My-short-c++-solution-O(1)-space-and-O(n)-time LeetCo...
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. 数组哈希法 复杂度 O(N) 时间 O(1) 空间 思路 这道题先得理解深刻,再做 对于...
First Missing Positive(Hard) Given an unsorted integer array, find the smallest missing positive integer. Example 1: Example 2: Example 3: Note: Your algorithm should run in O(...Leetcode 41. First Missing Positive Given an unsorted integer array, find the smallest missing positive integer....
First Missing Positive (java) Given an unsorted integer array, find the smallest missing positive integer. Note: Your algorithm should run in O(n) time and uses constant extra space. 难度:hard 题意:给一个未排序的整数数组,找到未出现的最小正整数。 思路:调整数组使每个值......
-231<= nums[i] <= 231- 1 题目大意:找到第一个缺失的正整数 解题思路:先讲数组放到map中,然后依次对比map中是否存在i,如果不存在就返回结果 classSolution(object):deffirstMissingPositive(self,nums):""":type nums: List[int]:rtype: int"""ifnums[0]==1andnums[len(nums)-1]==len(nums)andsum...
public int missingNumber(int[] nums) { int res = 0; for(int i = 0; i <= nums.length; i++){ res ^= i == nums.length ? i : i ^ nums[i]; } return res; } } First Missing Positive Given an unsorted integer array, find the first missing positive integer. ...
class Solution: # 本题直接用个for循环即可# 本题需要注意的是给定的未排序数组列表nums里面可能有重复数字,所以下面判断相邻两个数大小关系的的时候需要注意 def firstMissingPositive(self, nums): """ :type nums: List[int]-->给定的未排序数组列表:rtype: int-->final_min_number """ # 定义最终找到...
publicclassSolution{ publicintfirstMissingPositive(int[]nums){ if(nums==null||nums.length==0){ return1; } // 创建一个查找表,用来记录 1~nums.length 中数字出现的情况 boolean[]exist=newboolean[nums.length]; for(inti=0;i<nums.length;i++){ ...