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Lc0167

05 August 2008

*167 - 2Sum II - Input Array is Sorted

原题概述

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:

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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.

题意和分析

是上一道题1-2Sum的变种，按照第二种办法先排序再两个指针找的情况，因为input是sorted的数组，所以连排序都不用了，而且也不用一个额外的数组来存排序之前的indices，相比之下是更简单了。

Time：O(n)，Space：O(1)。

的当然，跟上一道题一样，也可以用HashMap来解，无论输入的array是否sorted,

Time: O(n), Space: O(n)

代码

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

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class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] result = new int[2];
        
        int left = 0, right = numbers.length - 1;
        while (left < right) {
            if (numbers[left] + numbers[right] < target) {
                left++;
            } else if (numbers[left] + numbers[right] > target) {
                right--;
            } else {
                break;//遇到合适的pair直接break，省点时间，有唯一解的情况
            }
        }
        
        //因为两个indices不是zero-based，所以+1，从1开始数
        result[0] = left + 1;
        result[1] = right + 1;
        
        return result;
    }
}
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