LeetCodeOJ——1.Tow Sum

Tow Sum

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].*

解决思路：两个循环，遍历。

代码：

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

后记：这是我最开始想到的方法，并且也AC了。然而时间复杂度为O（n²）。后来看到了别人的方法，利用map（以空间换时间），将时间复杂度降至O（n）。

代码如下：

两遍遍历：

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        map.put(nums[i], i);
    }
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement) && map.get(complement) != i) {
            return new int[] { i, map.get(complement) };
        }
    }
    throw new IllegalArgumentException("No two sum solution");
}

一遍遍历：

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution");
}