題目描述 - 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 解題
/**
* 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
- 首先扔茅，這里的正負法，邏輯是挨個取數(shù)組里的值序调，然后將對應(yīng)位置的數(shù)取負再扭；
- 之后再遍歷，位置是正的位置励背，就是沒出現(xiàn)的數(shù)；
- 組裝然后返回就可以了。
