leetcode 34. 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].

给定一个有序的数组和一个值,找到这个值在数组中的起始位置,如果找不到返回[-1, -1],要求时间复杂度在O(log n)之下.

时间复杂度为O(log n),则要用二分搜索了.

注意找值的开起位置和结束位置的二分搜索有所区别.

public class A34SearchforaRange {public int[] searchRange(int[] nums, int target) {int[] ans = {-1, -1};int i = 0;int j = nums.length - 1;ans[0] = binarySearchStart(nums, i, j, target);ans[1] = binarySearchEnd(nums, i, j, target);if(nums[ans[0]] != target || nums[ans[1]] != target)return new int[]{-1, -1};return ans;    }public int binarySearchStart(int[] nums, int i, int j, int target) {if(i < j) {int mid = i + (j - i) / 2;if(nums[mid] < target)return binarySearchStart(nums, mid + 1, j, target);else if(nums[mid] > target)return binarySearchStart(nums, i, mid - 1, target);elsereturn binarySearchStart(nums, i, mid, target);}return i;}public int binarySearchEnd(int[] nums, int i, int j, int target) {if(i < j) {int mid = i + (j - i) / 2 + 1;if(nums[mid] > target)return binarySearchEnd(nums, i, mid - 1, target);else if(nums[mid] < target)return binarySearchEnd(nums, mid + 1, j, target);elsereturn binarySearchEnd(nums, mid, j, target);}return i;}}