1. Two Sum

Problem statement:

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].

Solution one: hash table

This is a very classical question. It is asked most times in interview. According to the description, we need to return the indices of two numbers whose sum is equal to the target. By checking and inserting current number into a hash table, we can find the answer or for following searching.

Time complexity is O(n). Space complexity is O(n).

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

1. Two Sum

原文：http://www.cnblogs.com/wdw828/p/6881422.html