[LeetCode1] Two Sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

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

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

Analysis

1. Hash

从左往右扫描一遍，然后将数及坐标，存到map中。然后再扫描一遍即可。时间复杂度O(n)

2. 暴力搜索
这个比较省事，时间复杂度O(n*n)

java

public int[] twoSum(int[] numbers, int target) {
        int index[] = {0,0};
		HashMap<Integer, Integer> num = new HashMap<Integer,Integer>();
		for(int i=0;i<numbers.length;i++){
			num.put(numbers[i], i);
		}
		for(int i=0;i<numbers.length;i++){
			int temp = target-numbers[i];
			if(num.containsKey(temp)&&i<num.get(temp)){
				index[0] = i+1;
				index[1] = num.get(temp)+1;
				break;
			}
		}
		return index;
    }

c++

vector<int> twoSum(vector<int> &numbers, int target) {
     map<int,int> mapping;
    vector<int> result;
    for(int i=0;i<numbers.size();i++){
        mapping[numbers[i]] = i;
    }
    for(int i=0;i<numbers.size();i++){
        int searched = target-numbers[i];
        if(mapping.find(searched)!=mapping.end()){//find the element
            if(i<mapping[searched]){
                result.push_back(i+1);
                result.push_back(mapping[searched]+1);
                return result;
            }
            if(i>mapping[searched]){
                result.push_back(mapping[searched]+1);
                result.push_back(i+1);
                return result;
            }
        }
    }
    return result;
    }

solution2

java

public int[] twoSum(int[] numbers, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int index[] = {0,0};
		for(int i=0; i<numbers.length; i ++){
			for(int j=i+1; j<numbers.length; j++){
				if(target == numbers[i]+numbers[j]){
					index[0] = i+1;
					index[1] = j+1;
					return index;
				}
			}
		}
		return index;
    }