LeetCode 1.Two 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, and you may not use the sameelement twice.

Example:

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

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

/*
    思路：
    目标就是要找满足nums[current] + nums[another] == target，且current！=another的current和another
    为了快速的定位另一个加数，可以使用map。map的key是要查找的加数，value是该加数对应的位置。
    而map的确定和查找正好可以同步确定。故遍历一遍nums就可以解决问题。
     */
    public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            int another = target - nums[i];

            Integer indice = map.get(another);
            if (indice != null) {
                if (indice != i) {
                    result[0] = i;
                    result[1] = indice;
                    return result;
                }
            }else{
                map.put(nums[i], i);
            }
        }
        return result;
    }
//    暴力方法
//    public int[] twoSum(int[] nums, int target) {
//        int[] result = new int[2];
//        for (int low = 0; low < nums.length - 1; low++) {
//            for (int high = nums.length - 1; high > low; high--) {
//                if (nums[low] + nums[high] == target){
//                    result[0] = low;
//                    result[1] = high;
//                    return result;
//                }
//            }
//        }
//        return result;
//    }