448. Find All Numbers Disappeared in an Array

问题描述：

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]

解题思路：

首先计算出数组中出现的数字x的下标形式，则下表为index = |x| - 1，可将数组a中得a[index]标为负数来标记数字x出现在数组中。然后若数组中某数字已为负数，说明该数字在数组中出现过。最终遍历数组，若a[i]为正数，表明数字 i + 1在数组中没有出现过。

#include"stdio.h"#include<stdlib.h>void findDisappearedNumbers(int *a,int n){    int i,index;    int b[n],j=0;    for(i=0;i<n;i++){        index=abs(a[i])-1;        if(a[index]>0)        a[index]*=(-1);
    }    for(i=0;i<n;i++){        if(a[i]>0)        {            b[j]=i+1;j++;}    }
    for(i=0;i<j;i++){        printf("%d\n",b[i]);    }}
void main(){    int q[8]={4,3,2,7,8,2,3,1},i;
    findDisappearedNumbers(q,8);
}