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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].
题目大意：给一个排序后的数组，和一个target value，找到这个数字在数组中的开始位置和结束位置（可能有多个），如果没有找到target，返回-1，-1
复杂度必须为logN级
思路：logn必为二分查找，找到第一个等于target的element，向左和向右找到range的起始
算法构架：增加一个二分查找函数，返回找到的第一个等于target的element的位置，然后从当前位置向前和向后找到边界
public class Solution { public int[] searchRange(int[] nums, int target) { int[] res = new int[] {-1,-1}; if (nums == null || nums.length == 0) { return res; } int start = binarySearch(nums, target); int temp = start; if (start == -1) { return res; } while (temp >= 0 && nums[temp] == target) { temp--; } res[0] = temp+1; temp = start; while (temp < nums.length && nums[temp] == target) { temp++; } res[1] = temp-1; return res; } private int binarySearch(int[] nums, int target) { int left = 0; int right = nums.length-1; while (left <= right) { int mid = left + (right-left) / 2; if (nums[mid] == target) { return mid; } else if (nums[mid] > target) { right = mid-1; } else { left = mid+1; } } return -1; } }
LeetCode: 34. Search for a Range
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原文地址：http://www.cnblogs.com/snakech/p/5760476.html
