题目描述 - leetcode

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]

C 解题

/**
 * Return an array of size *returnSize.
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* findDisappearedNumbers(int* nums, int numsSize, int* returnSize) {
    
    for(int i=0;i<numsSize;i++){
        int index=abs(nums[i]);
        nums[index-1]=-abs(nums[index-1]);
    }
    
    int count = 0;
    for(int i=0;i<numsSize;i++){
        if(nums[i]>0){
            count += 1;
        }
    }
    
    int* a = malloc (count * sizeof(int));
    *returnSize = count; 
    
    count = 0;
    for(int i=0;i<numsSize;i++){
        if(nums[i]>0){
            a[count] = i+1;
            count += 1;
        }
    }
    
    return a;
}

分析

• 首先，这里的正负法，逻辑是挨个取数组里的值，然后将对应位置的数取负；
• 之后再遍历，位置是正的位置，就是没出现的数；
• 组装然后返回就可以了。