Description:

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].

Discuss:

1.暴力法:

时间复杂度: O(n^2)

空间复杂度: O(1)

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int res[] = {0,0};
        for (int i = 0;i<nums.length;++i){
            for (int j = i+1;j<nums.length;++j){
                if (target == (nums[i] + nums[j])){
                    res[0] = i;
                    res[1] = j;
                    }
                }
            }
        return res;
    }
}
class Solution {
    public int[] twoSum(int[] nums, int target) {
        for (int i = 0;i<nums.length;++i){
            for (int j = i+1;j<nums.length;++j){
                if (target == (nums[i] + nums[j]))
                    return new int[]{i,j};
            }
        }
        return null;
    }
}

2.Hash表

时间复杂度: O(n)

空间复杂度: O(n)

思路:利用哈希表将数组中的值依次存入，并且在数组遍历过程中找到另一个值

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer  map = new HashMap< ();
            for (int i = 0; i < nums.length; i++){
                if(map.containsKey(target - nums[i])){
                    int[] r = new int[]{map.get(target - nums[i]), i} ;
                    return r;
                }else{
                map.put(nums[i], i);    
            }
        }
        return null;        
    }
}