167. Two Sum II - Input array is sorted.md

167. Two Sum II - Input array is sorted

Question

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.

Thinking:

• Method 1: O(N^2)

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int len = numbers.length;
        for(int i = 0; i < len-1; i++){
            int find = target - numbers[i];
            for(int j = i + 1; j < len; j++){
                if(numbers[j] == find){
                    return new int[]{i + 1, j + 1};
                }else if(numbers[j] > find)
                    break;
            }
        }
        return null;
    }
}

• Method 2: O(N)
  • 因为数组本来是按序排列的，我们可以利用sum 3的方法。
    • 求出首位的和，如果小了，左边的右移。
    • 如果打了，右边的左移。

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int[] result = new int[2];
        int low = 0, high = numbers.length - 1;
        if(numbers.length < 2) return null;
        while(low < high){
            int sum = numbers[low] + numbers[high];
            if(sum == target){
                result[0] = low + 1;
                result[1] = high + 1;
                break;
            }else if(sum > target)
                high--;
            else
                low++;
        }
        return low >= high ? null : result;
    }
}

二刷

1. 双指针，一个指向数组头，一个指向数组尾。
2. 如果sum小了，low指针右移，不然high指针左移。

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int len = numbers.length;
        int low = 0, high = len - 1;
        int sum = numbers[low] + numbers[high];
        while(sum != target){
            if(sum < target) low++;
            else high--;
            sum = numbers[low] + numbers[high];
        }
        return new int[]{low + 1, high + 1};
    }
}

Third Time

• Method 1: Two pointer

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