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. 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 and you may not use the same element twice.

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

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二分查找方法

public int[] twoSum(int[] numbers, int target) {
        int N = numbers.length;
		for(int i = 0; i < N; i++)
		{
			int start = i + 1;
			int end = N - 1;
			int diff = target - numbers[i];
			while(start <= end)
			{
				int mid = start + (end - start)/2;
				if(numbers[mid] == diff)
					return new int[]{i+1,mid+1};
				else if(numbers[mid] > diff)
					end = mid - 1;
				else
					start = mid + 1;
			}
		}
		return new int[]{};
    }

直接查找

使用两个指针分别指向数组的首尾

public int[] twoSum(int[] numbers, int target) {
        int i = 0;
        int j = numbers.length - 1;
        while(i < j)
        {
           int tmp = numbers[i] + numbers[j];
            if(target == tmp)
                return new int[]{i+1,j+1};
            if(target > tmp)
                i += 1;
            else
                j -= 1;
        }
        return new int[]{};
    }

这道题目比较简单，我觉的需要注意的有两点

• 函数最后的返回值，return

• 根据Target的值来进行指针的移动