leetcode problem 41 -- 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.

题意:

　　给定了一个无序数组,要求找到第一个缺失的正整数.如1,2,4,5的第一个缺失的正整数是3.要求时间复杂度为o(n) 空间复杂度为o(1)

思路:

　　第一种是排序,时间复杂度为O(nlgn) 不行

　　第二种是hash查找对每一个数字都用额外空间找到正确的位置. 然后一次遍历找到第一个缺失的正整数. 时间复杂度为O(n) 但是空间复杂度为O(n) 也不符合题目要求

　　第三种直接在输入数组上进行交换操作, 使得每个数字通过交换能在正确的位置上.如3,1,7,2 这组数字第一次交换后得到7,1,3,2 数字3放置在了正确的位置.

　　　　这种时间复杂度为O(n) 而且只需要常数辅助空间. 但我认为这种方法依然不复合题目要求,因为它破坏了原始数组. 它其实也类似于hash, 需要O(n)的空间,只不过这O(n)的空间借用了原始输入数组上的.

　　目前我还没找到即不破坏原始数组,空间复杂度又为O(1)的方法

代码1(hash法):

class Solution {
public:int firstMissingPositive(vector<int>& nums) {if (nums.size() == 0)return 1;auto it_max = std::max_element(nums.begin(), nums.end());    auto it_min = std::min_element(nums.begin(), nums.end());vector<bool> vec(*it_max - *it_min, false);for (auto elem : nums) vec[elem - *it_min] = true;for (auto it = 1; it <= *it_max; ++it) {if (it < *it_min || (it > 0 && !vec[it - *it_min]))return it;}    return *it_max > 0 ? *it_max + 1 : 1;}private:
};

代码2(直接在原数组上进行交换, 空间代价为o(1))

class Solution {
public:int firstMissingPositive(vector<int>& nums) {if (nums.empty())return 1;auto begin = getFirstPositivePos(nums);if (*begin < 0)return 1;for (auto it = begin; it != nums.end(); ) {auto realPos = begin + *it - 1;if (realPos < nums.end() && realPos != it && *realPos != *it)iter_swap(it, realPos);else++it;}int index = 1;for (auto it = begin; it != nums.end(); ++it, index++) {if (index != *it)return index;}return index;}private:vector<int>::iterator getFirstPositivePos(vector<int>& nums) {auto first = nums.begin();auto last = nums.end()-1;while (true) {while (first < last && *first <= 0 ) ++first;while (first < last && *last > 0)--last;if (first < last)iter_swap(first, last);    elsebreak;}return first;}
};

转载于:https://www.cnblogs.com/lysuns/p/4479212.html

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