1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

从给定序列中找出和为指定数字的两个数字的下标

You may assume that each input would have exactly one solution.

假设每个输入只有一个解

Example:

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

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

Brute Force

time complexity O(n2)
space complexity O(1)
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class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> result;
        int len = nums.size();
        for(int i = 0; i < len - 1; ++i){
            for(int j = i + 1; j < len; ++j){
                if(nums[i]+nums[j] == target){
                    result.push_back(i);
                    result.push_back(j);
                    return result;
                }
            }
        }
        return result;
    }
};

Hash

time complexity O(n)
space complexity O(n)
该方法给出的序列是从后往前的
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class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    unordered_map<int, int> imap;
    for (int i = 0;; ++i) {
        auto it = imap.find(target - nums[i]);        
        if (it != imap.end()) 
            return vector<int> {i, it->second};            
        imap[nums[i]] = i;
        }
    }
};








posted @ 2016-11-26 13:29  copperface  阅读(565)  评论(0)  编辑  收藏  举报