1. Two Sum

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题目

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].

解题思路

  • 暴力破解，遍历每个元素x并查找是否有另一个值等于target - x。
  • 时间复杂度为 $O(n^2)$

代码

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

更优解法

  • 这个是看了答案发现的，利用map的查找，节省时间。
  • 时间复杂度为 $O(nlogn)$
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class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> ans;
        map<int, int> myMap;
        for(int i = 0; i < nums.size(); ++i) {
            if(myMap.count(target - nums[i])) {
                ans.push_back(myMap.find(target - nums[i])->second);
                ans.push_back(i);
                return ans;
            } 
            else
                myMap.insert(make_pair(nums[i], i));
        }
    }
};