【leetcode】1. Two Sum

一、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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

UPDATE (2016/2/13):

The return format had been changed to zero-based indices. Please read the above updated description carefully.

题目解读：给一个整数序列和一个目标值，找出序列中两个数加起来等于目标值，返回这两个数在序列中的下标

思路：一开始只能想到写两个for循环遍历

c++代码（758ms，11.48%） 就这样还能排到11%呢  = = ！ 汗....

class Solution {
public:
vector<int> twoSum(vector<int>& nums, int target) {
vector<int> result;
int left,right;
for(left=0; left<nums.size(); left++){
for(right = left+1; right<nums.size(); right++){
if((nums[left]+nums[right]) == target){
result.push_back(left);
result.push_back(right);
return result;
}//if
}//for
}//for
return result;
}
};

看讨论区别人的代码，在遍历的同时使用map存储已遍历过的数的位置，以便于快速查找，时间复杂度可以达到o(n)

c++代码（16ms，62.06%）

class Solution {
public:
vector<int> twoSum(vector<int>& numbers, int target) {
//Key is the number and value is its index in the vector.
unordered_map<int, int> hash;
vector<int> result;
for (int i = 0; i < numbers.size(); i++) {
int numberToFind = target - numbers[i];

//if numberToFind is found in map, return them
if (hash.find(numberToFind) != hash.end()) {
//+1 because indices are NOT zero based
result.push_back(hash[numberToFind]);
result.push_back(i);
return result;
}

//number was not found. Put it in the map.
hash[numbers[i]] = i;
}
return result;
}
};