167. Two Sum II - Input array is sorted

Leetcode Array Two Pointers Binary Search

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.

分析

一种最简单的方法和LeetCode 1 Two Sum一样，利用哈希表来处理。代码是一摸一样的。

Java

public static int[] twoSum(int[] nums, int target) {
    if (nums == null || nums.length < 2) return new int[]{};
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i< nums.length; i++)
        if (map.containsKey(nums[i]))
            return new int[]{map.get(nums[i]), i};
        else map.put(target - nums[i], i); 
    return new int[]{};
}

另一种方法是利用两个指针，慢慢逼近所求值: 先在数组中选择两个数字

• 如果它们的和等于输⼊的target，我们就找到了要找的两个数字。
• 如果它们的和小于输入的target，为了减小和，可以选择前⾯的数字，因为排在数组前⾯的数字要⼩⼀些。
• 如果它们的和大于输入的target，为了增加和，可以选择较⼩的数字后⾯的数字，因为排在后⾯的数字要⼤⼀些。

public int[] twoSumTwoPointers(int[] nums, int target) {
    int left = 0, right = nums - 1;
    while (nums[left] + nums[right] != target)
        if (nums[left] + nums[right] > target) right--;
        else left++;
    return new int[]{left + 1, right + 1};
}