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.

Example:

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

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

解题思路

我想大多数人和我一样都会想到用这种方法。写两个for循环，依次查找看A[i] + A[j] 是否等于target.若等于返回i，j.

思路2：对于java 我们可以采用Map（以空间换时间）时间复杂度为O(n)。

通过map来查找a和target-a是不是都在数组中,如果在则返回他们的下标。

代码：

import java.util.*;
public class Solution {
//思路1
    public int[] twoSum(int[] nums, int target) {
        int[] A = new int[2];
        A[0] = A[1] = -1;
        for(int i = 0; i<nums.length-1;i++)
        for(int j = i+1 ;j<nums.length;j++)
        {
            if((nums[i] + nums[j] ) == target)
            {
                A[0] = i;
                A[1] = j;
            }
        }
        return A;
    }
//思路2
public int[] twoSum(int[] numbers, int target) {
    int[] result = new int[2];
    Map<Integer, Integer> map = new HashMap<Integer, Integer>();
    for (int i = 0; i < numbers.length; i++) {
        if (map.containsKey(target - numbers[i])) {
            result[1] = i ;
            result[0] = map.get(target - numbers[i]);
            return result;
        }
        map.put(numbers[i], i );
    }
    return result;
}
}