167 Two Sum II - Input array is sorted-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_33669968/article/details/93545639

博客围绕有序整数数组，求解两数之和等于特定目标值的问题。介绍了两种解法，一是游标滑动并查找，时间复杂度O(NlogN)，空间复杂度O(1)；二是指针对撞，两指针从两端向中间逼近，时间复杂度O(N)，空间复杂度O(1)。

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题目

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 same element 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.

解法思路（一）

游标 i 在 [0, numbers.length - 1) 中滑动的过程中，在 [i + 1, numbers.length-1] 中找 target - numbers[i]；

解法实现（一）

时间复杂度

O(NlogN)；

空间复杂度

O(1)；

关键词

二分查找

package leetcode._167;

/**
 * 从 i 开始遍历，然后在剩下的数组中用二分查找法找 target - numbers[i]；
 * 时间复杂度：O(NlogN)
 */
public class Solution2 {

    public int[] twoSum(int[] numbers, int target) {
        for(int i = 0; i < numbers.length - 1; i++) {
            int j = binarySearch(numbers, i + 1, numbers.length-1, target - numbers[i]);
            if (j != -1) {
                return new int[] {i+1, j+1};
            }
        }
        return null;
    }

    private int binarySearch(int[] numbers, int l, int r, int target) {
        while (l <= r) {
            int mid = (l + r)/2;
            if (numbers[mid] == target) {
                return mid;
            } else if (numbers[mid] < target) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        int[] numbers = {2, 7, 11, 15};
        int[] result = (new Solution2()).twoSum(numbers, 9);
        for(int i = 0; i < result.length; i++) {
            System.out.println(result[i]);
        }
    }

}

算法思路（二）

指针对撞；
两个指针 i 和 j 分别从数组的两端向中间逼近，比较两个指针所指元素之和与 target 的大小；

解法实现（二）

时间复杂度

O(N)；

空间复杂度

O(1)；

关键词

对撞指针

package leetcode._167;

/**
 * 对撞指针
 * 时间复杂度：O(N)
 * 空间复杂度：O(1)
 */
public class Solution3 {

    public int[] twoSum(int[] numbers, int target) {

        int i = 0, r = numbers.length - 1;

        while (i < r) {
            if (numbers[i] + numbers[r] == target) {
                return new int[]{i + 1, r + 1};
            } else if (numbers[i] + numbers[r] < target) {
                i ++;
            } else {
                r --;
            }
        }

        return null;
    }

    public static void main(String[] args) {
        int[] numbers = {2, 7, 11, 15};
        int[] result = (new Solution3()).twoSum(numbers, 9);
        for(int i = 0; i < result.length; i++) {
            System.out.println(result[i]);
        }
    }

}