2341. How many pairs can an array form?
You are given an integer array nums with indexes starting from 0. In one step, you can perform the following steps:
从 nums 选出 两个 相等的 整数
从 nums 中移除这两个整数，形成一个 数对
Please perform this operation on nums multiple times until it cannot continue.
Returns an integer array answer starting from 0 and of length 2 as the answer, where answer[0] is the number of pairs formed and answer[1] is the number of integers remaining after performing as many of the above operations on nums as possible.
Example 1:
Input: nums = [1,3,2,1,3,2,2]
Output: [3,1]
Explanation:
nums[0] and nums[3] form a number pair and are removed from nums, nums = [3,2,3,2,2] .
nums[0] and nums[2] form a number pair and are removed from nums, nums = [2,2,2].
nums[0] and nums[1] form a number pair and are removed from nums, nums = [2].
No more pairs can be formed. A total of 3 number pairs are formed, leaving 1 number in nums.
Example 2:
Input: nums = [1,1]
Output: [1,0]
Explanation: nums[0] and nums[1] form a number pair and are removed from nums, nums = [].
No more pairs can be formed. A total of 1 number pair is formed, leaving 0 numbers in nums.
Example 3:
Input: nums = [0]
Output: [0,1]
Explanation: Unable to form a number pair, there is 1 number left in nums.
This question is relatively simple. The solution code is as follows:
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* numberOfPairs(int* nums, int numsSize, int* returnSize){
int count[101];
int * re=(int *)malloc(sizeof(int)*2);
*returnSize=2;
for(int i=0;i<101;i++){
count[i]=0;
}
re[0]=0;
re[1]=0;
for(int i=0;i<numsSize;i++){
count[nums[i]]++;
if(count[nums[i]]==2){
re[0]++;
re[1]--;
count[nums[i]]=0;
}
if(count[nums[i]]==1){
re[1]++;
}
}
return re;
}