## 题目地址

https://leetcode.com/problems/two-sum

## 题目描述

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].
``````

## 代码

C++ 解法一：

``````class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> m;
vector<int> res;
for (int i = 0; i < nums.size(); ++i) {
m[nums[i]] = i;
}
for (int i = 0; i < nums.size(); ++i) {
int t = target - nums[i];
if (m.count(t) && m[t] != i) {
res.push_back(i);
res.push_back(m[t]);
break;
}
}
return res;
}
};

``````

Java 解法一：

``````public class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
int[] res = new int[2];
for (int i = 0; i < nums.length; ++i) {
m.put(nums[i], i);
}
for (int i = 0; i < nums.length; ++i) {
int t = target - nums[i];
if (m.containsKey(t) && m.get(t) != i) {
res[0] = i;
res[1] = m.get(t);
break;
}
}
return res;
}
}

``````

C++ 解法二：

``````class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
unordered_map<int, int> m;
for (int i = 0; i < nums.size(); ++i) {
if (m.count(target - nums[i])) {
return {i, m[target - nums[i]]};
}
m[nums[i]] = i;
}
return {};
}
};

``````

Java 解法二：

``````public class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
int[] res = new int[2];
for (int i = 0; i < nums.length; ++i) {
if (m.containsKey(target - nums[i])) {
res[0] = i;
res[1] = m.get(target - nums[i]);
break;
}
m.put(nums[i], i);
}
return res;
}
}
``````

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