leetCode 167. Two Sum II

Given an array of integers that is already sorted in ascending order, 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.

Note:

• Your returned answers (both index1 and index2) are not zero-based.
• You may assume that each input would have exactly one solution and you may not use the sameelement twice.

Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

class Solution5 {
//  对于有序的数组这种方法会快很多
    /*
    这种方法的合理性：
        假设numbers[low']+numbers[high'] = target    -------①
        对于有序的整数数组，low只能++，high只能--，总会出现low走到low'，或者high走到high'的情况。
        情况一：
            low先走到low'：此时numbers[low]+numbers[high] > target, 只有high会移动，而low不变。一直等到①式成立为止。
        情况二：
            high先走到high'：此时numbers[low]+numbers[high] < target, 只有low会移动，而high不变。一直等到①式成立为止。

     */
    public int[] twoSum(int[] numbers, int target) {
        int[] result = new int[2];
        int low = 0, high = numbers.length - 1;
        while (low < high){
            int sum = numbers[low] + numbers[high];
            System.out.println(numbers[low] + "+"+numbers[high]+" = "+sum);
            if (sum > target){
                --high;
            }else if (sum < target){
                ++low;
            }else {
                return new int[]{low+1, high+1};
            }
        }
        return result;
    }

//    普通方法，有序无序都可以
    public int[] twoSum(int[] numbers, int target) {
        int[] result = new int[2];
        //key：查找的数字，value：该数字对应的位置
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = numbers.length - 1; i >= 0; i--) {
            /*
            检验map中是否有和 i 位置的数的和为target的，有结束查找，没有将i位置的数及其位置存入map
             */
            Integer indiceOfAnother = map.get(target - numbers[i]);
            if (indiceOfAnother == null){
                map.put(numbers[i],i);
            }else {
                result[0] = i+1;
                result[1] = indiceOfAnother+1;
                return result;
            }
        }
        return result;
    }
}