1.描述
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.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
2.分析
3.代码
C语言版
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* nums, int numsSize, int target) {
for (int i = 0; i < numsSize; ++i) {
for (int j= i + 1; j < numsSize; ++j) {
if (nums[i] + nums[j] == target) {
int* ans = (int*)calloc(2,sizeof(int));
ans[0] = i;
ans[1] = j;
return ans;
}
}
}
return NULL;
}
C++版
class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> v;
for (int i = 0; i < nums.size(); ++i) {
for (int j = i + 1; j < nums.size(); ++j) {
if (nums[i] + nums[j] == target) {
v.push_back(i);
v.push_back(j);
return v;
}
}
}
return v;
}
};
