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...
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...
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. 寻找第一个miss的正数,把数组排序就ok了,因为空间复杂度需要为1,所以在原数组上对元素交换就行。
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......
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 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.
-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...
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....
First Missing Positive 同样是一道我不太能理解为什么能标为hard的题目。 我的解法是将所有正数都先放到map里面,然后就从小正数——也就是1——开始检查map,遇到的第一个不包含在map中的正数便是答案。最坏情况下的复杂度是O(n)。 javapublic class Solution { ...
class Solution { public: int firstMissingPositive(vector<int>& nums) { int n = nums.size(); for (int i = 0; i < n; ++i) { // 将nums[i]放到正确的位置,直到无法交换为止 while (nums[i] >= 1 && nums[i] <= n && nums[nums[i] - 1] != nums[i]) { swap(nums[i], nums...