Leetcode/ sum of two numbers in a sorted array

Code

package com.xcrj;

import java.util.Arrays;

/** *  The finger of the sword  Offer II 006.  Sort the sum of two numbers in the array  *  Given a has been according to   Ascending order   Array of integers for numbers , Please find out two numbers from the array, and the sum of them is equal to the target number target . *  Functions should be length based  2  Returns the subscript values of the two numbers in the form of an array of integers .numbers  The subscript   from  0 Start counting  , So the answer array should satisfy  0<= answer[0] < answer[1] <numbers.length. *  Suppose that there is only one pair of qualified numbers in the array , At the same time, a number cannot be used twice . */
public class Solution6 {
    
    /** * target-numbers[i] Do a binary search ( Binary search ) *  Premise , Ordered sequence , It is known that numbers[] Array sequence order  */
    public int[] twoSum1(int[] numbers, int target) {
    
        for (int i = 0; i < numbers.length; i++) {
    
            int idx = binarySearch(numbers, target - numbers[i], i + 1, numbers.length - 1);
            //  Binary search found the result 
            if (-1 != idx) {
    
                return new int[]{
    i, idx};
            }
        }

        return new int[0];
    }

    /** *  Two points search （ Binary search ） *  The target value is equal to the intermediate value and returns the index  *  The target value is greater than the middle value. Look for the right half  *  The target value is less than the middle value. Look for the left half  */
    public int binarySearch(int[] numbers, int x, int start, int end) {
    
        int i = start;
        int j = end;
        while (i <= j) {
    
            int mid = (i + j) / 2;
            if (x == numbers[mid]) return mid;
            if (x > numbers[mid]) i = mid + 1;
            else j = mid - 1;
        }

        return -1;
    }

    /** *  Known ascending sequence  *  Double pointer  i j  Move in the opposite direction , Each move 1 A pointer to the  */
    public int[] twoSum2(int[] numbers, int target) {
    
        int i = 0;
        int j = numbers.length - 1;
        while (i < j) {
    
            int sum = numbers[i] + numbers[j];
            if (sum == target) {
    
                return new int[]{
    i, j};
            }
            //  Less than i Move right 
            if (sum < target) i++;
                //  Greater than j Move left 
            else {
    
                j--;
            }
        }

        return new int[0];
    }

    public static void main(String[] args) {
    
        Solution6 solution6 = new Solution6();
        System.out.println(Arrays.toString(solution6.twoSum2(new int[]{
    1, 2, 4, 6, 10}, 8)));
    }
}

Reference resources

author ：tangweiqun
link ：https://leetcode.cn/problems/kLl5u1/solution/jian-dan-yi-dong-javac-pythonjs-liang-sh-et4y/
source ： Power button （LeetCode）