LeetCode——Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

原题链接:https://oj.leetcode.com/problems/search-for-a-range/

题目：给定一个排好序的整数数组，找到给定目标值的出现的首尾位置。

思路：二分查找。由于是有序数组，所以相同值的数是连续的，即只要找到其中一个，再向左右找到边界值就可以了，这三步均采用二分查找。

public int[] searchRange(int[] A, int target) {int result[] = new int[] { -1, -1 };int len = A.length;if (len <= 0)return result;int separator = -1;int left = 0, right = len - 1;while (left <= right) {int middle = (left + right) / 2;if (A[middle] > target)right = middle - 1;else if (A[middle] < target)left = middle + 1;else {separator = middle;break;}}if (separator == -1)return result;//找到左边界left = 0;right = separator;while (left <= right) {int middle = (left + right) / 2;if (A[middle] == target)right = middle - 1;elseleft = middle + 1;}result[0] = left;//找到右边界left = separator;right = len - 1;while (left <= right) {int middle = (left + right) / 2;if (A[middle] == target)left = middle + 1;elseright = middle - 1;}result[1] = right;return result;}

reference : http://www.darrensunny.me/leetcode-search-for-a-range/