NO.5_448. Find All Numbers Disappeared in an Array

448. Find All Numbers Disappeared in an Array

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

Input:[4,3,2,7,8,2,3,1]Output:[5,6]

//参考解

class Solution {public:vector<int> findDisappearedNumbers(vector<int>& nums) {    vector<int> res;    if(nums.empty()) return res;    int n = nums.size();    for(int i = 0; i < n; ++i){        while(nums[nums[i]-1] != nums[i]){            swap(nums[nums[i]-1], nums[i]);        }    }    for(int i = 0; i < n; ++i){        if(nums[i] != i+1){            res.push_back(i+1);        }    }    return res;}};

//解1：超时

class Solution {public:    vector<int> findDisappearedNumbers(vector<int>& nums)     {        vector<int>a;        int data_size=nums.size();        a.resize(data_size);                for(int i=1;i<=data_size;i++)//a初始化        {            a[i]=i;        }                for(int i=0;i<nums.size();i++)//用0占位        {            a[nums[i]]=0;        }               a.erase(remove(a.begin(),a.end(),0),a.end());//移除vector所有的指定元素                 return a;    }};

//解2

class Solution {public:vector<int> findDisappearedNumbers(vector<int>& nums) {    vector<int> res;    int n = nums.size();        for(int i = 0; i < n; ++i)    {        if(nums[i]==nums[nums[i]-1])continue;//这个元素（nums[i]）的位置（nums[nums[i]-1]）上，正好是这个元素（原因：nums[i]数量不唯一）        do        {           swap((nums[i]),nums[nums[i]-1]);//把这个元素（nums[i]）换到属于它的位置（nums[nums[i]-1]）上，位置（i）上换来了一个新元素        }        while(nums[i]!=nums[nums[i]-1]);//直到位置（i）上换来了的某个新元素（nums[i]）的位置（nums[nums[i]-1]）上是它（nums[i]）    }    for(int i = 0; i < n; ++i){        if(nums[i] != i+1)        {            res.push_back(i+1);        }    }    return res;}};