给定一个整数数组和一个目标值,找出数组中和为目标值的两个数
你可以假设每个输入只对应一种答案,且同样的元素不能被重复使用
示例:
给定nums = [2,7,11,15],target = 9
因为nums[0]+nums[1] = 2 + 7 = 9
所以返回 [0,1]
c代码实现
#include <stdlib.h>
struct object
{
int val;
int index;
};
static int compare(const void* a, const void* b)
{
return ((struct object*)a)->val - ((struct object*)b)->val;
}
static int* twoSum(int* nums, int numSize, int target)
{
struct object* objs = (struct object*)malloc(numSize*sizeof(object));
for (int i = 0; i < numSize; ++i)
{
struct object& obj = objs[i];
obj.val = nums[i];
obj.index = i;
}
qsort(objs,numSize,sizeof(*objs),compare);
int i = 0;
int j = numSize - 1;
int* results = (int*)malloc(2*(sizeof(int)));
while (i < j)
{
int diff = target - objs[i].val;
if (diff > objs[j].val)
while (++i < j && objs[i].val == objs[i - 1].val) {}
else if(diff < objs[j].val)
while (--j > i && objs[j].val == objs[j + 1].val) {}
else
{
results[0] = objs[i].index;
results[1] = objs[j].index;
return results;
}
}
free(objs);
objs = NULL;
return NULL;
}
int main()
{
int arr[] = {3,4,5,6,7};
int* results = twoSum(arr, sizeof(arr) / sizeof(*arr), 11);
if (results)
{
printf("%d\n", arr[results[0]]);
printf("%d\n", arr[results[1]]);
free(results);
results = NULL;
}
else
{
printf("not found \n");
}
system("pause");
return 0;
}
c++代码实现
#include <vector>
#include <map>
vector<int> twoSum(int* nums, int numSize, int target)
{
map<int, int> m;
vector<int> vec;
for (int i = 0; i < numSize; ++i)
{
map<int, int>::iterator iter = m.find(target - nums[i]);
if (iter == m.end())
{
m[nums[i]] = i;
}
else
{
vec.push_back(iter->second);
vec.push_back(i);
return vec;
}
}
return vec;
}
int main()
{
int arr[] = {3,4,5,6,7};
vector<int> results = twoSum(arr, sizeof(arr) / sizeof(*arr), 10);
if (results.size()> 0)
{
printf("%d\n", arr[results[0]]);
printf("%d\n", arr[results[1]]);
}
else
{
printf("not found \n");
}
system("pause");
return 0;
}