Leetcode习题记录——Two Sum

题目：Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

方法一：
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* nums, int numsSize, int target) {
int *a = (int*)malloc(2*sizeof(int));
for(int i = 0 ; i < numsSize ; i++ ){
for (int j = i + 1 ; j < numsSize ; j++){
if( nums[i] + nums[j] == target){
a[0] = nums[i];
a[1] = nums[j];
}
}
}
return a;
}

这里是第一种对简单的解法，两次循环遍历所有的组合。思路很简单，值得一提的是这个malloc的用法：malloc (size_type size)

返回的必须是一个指针，此函数包含于 stdlib.h

方法二：

int compare(const void *a , const void *b){
return *(int*) a - *(int*) b;
}

int* twoSum(int* nums, int numsSize, int target) {
bool flag = true;
int* result = malloc(sizeof(int) * 2);
int* temp = malloc(sizeof(int) * numsSize);
qsort(nums,numsSize,sizeof(int),compare);
int low = 0;
int high = numsSize-1;
for (int i = 0; i < numsSize; i++){
if(nums[low] + nums[high] > target)
high--;
if(nums[low] + nums[high] < target)
low++;
if(nums[low] + nums[high] == target)
break;
}
result[0] = nums[low];
result[1] = nums[high];
return result;
}

这里是第二种做法，思路很简单，先将数组排序，设置两个指针一个指向最小值，一个指向最大值，将这两个数相加，如果超过target，那么说明大的数太大，所以最高的指针向下移动一个，指向一个稍微小一点的数，如果低于target，同理。排序的函数也不用自己写，在库中有，qsort( void *base, size_t num, size_t width, int (cdecl *compare )如何使用这个函数具体详见：http://blog.csdn.net/yuanjilai/article/details/7348345