【LeetCode】167.Two Sum II

【LeetCode】167.Two Sum II - Input array is sorted解题报告

tags: Array

题目地址：https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/#/description
题目描述：

  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. 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.

Example：

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

Solutions:

这里没有用hashtable，而是因为已经相对有序，就从两边向中间缩小范围。

public class Solution {    public int[] twoSum(int[] numbers, int target) {        int[] index = new int[2];        if (numbers == null || numbers.length < 2) return index;        int left = 0, right = numbers.length - 1;        while (left < right) {            long v = numbers[left] + numbers[right];            if (v == target) {                index[0] = left + 1;                index[1] = right + 1;                break;            } else if (v > target) {                right --;            } else {                left ++;            }        }        return index;    }}

因为既然是相对有序了，我们可以继续优化，减少时间复杂度，下面是更简单时间复杂度的方法：

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

Date：2017年6月3日