167. Two Sum II - Input array is sorted

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. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

链接： http://leetcode.com/problems/two-sum-ii-input-array-is-sorted/

题解：

排序好的数组求two sum。用头尾两个指针对着夹逼一下就可以了。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] res = {-1, -1};
        if(numbers == null || numbers.length == 0)
            return res;
        int lo = 0, hi = numbers.length - 1;
        
        while(lo < hi) {
            if(numbers[lo] + numbers[hi] < target)
                lo++;
            else if(numbers[lo] + numbers[hi] > target)
                hi--;
            else {
                res[0] = lo + 1;
                res[1] = hi + 1;
                return res;
            }
        }
        
        return res;
    }
}

 

二刷：

和一刷一样，双指针夹逼

Java:

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] res = {-1, -1};
        if (numbers == null || numbers.length == 0) {
            return res;
        }
        int lo = 0, hi = numbers.length - 1;
        while (lo <= hi) {
            int sum = numbers[lo] + numbers[hi];
            if (sum < target) {
                lo++;
            } else if (sum > target) {
                hi--;
            } else {
                res[0] = lo + 1;
                res[1] = hi + 1;
                return res;
            }
        }
        return res;
    }
}

 

三刷:

Java:

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        if (nums == null || nums.length < 2) return nums;
        int lo = 0, hi = nums.length - 1;
        while (lo < hi) {
            int sum = nums[lo] + nums[hi];
            if (sum == target) return new int[] {lo + 1, hi + 1};
            else if (sum < target) lo++;
            else if (sum > target) hi--;
        }
        return new int[] {-1, -1};
    }
}

 

 

 

Reference：