# Search for a Range ### Source - leetcode: [Search for a Range | LeetCode OJ](https://leetcode.com/problems/search-for-a-range/) - lintcode: [(61) Search for a Range](http://www.lintcode.com/en/problem/search-for-a-range/) ### Problem Given a sorted array of *n* integers, find the starting and ending position ofa given target value. If the target is not found in the array, return `[-1, -1]`. #### Example Given `[5, 7, 7, 8, 8, 10]` and target value `8`, return `[3, 4]`. #### Challenge O(log *n*) time. ### 题解 lower/upper bound 的结合，做两次搜索即可。 ### Java ~~~ public class Solution { /** *@param A : an integer sorted array *@param target : an integer to be inserted *return : a list of length 2, [index1, index2] */ public int[] searchRange(int[] A, int target) { int[] result = new int[]{-1, -1}; if (A == null || A.length == 0) return result; int lb = -1, ub = A.length; // lower bound while (lb + 1 < ub) { int mid = lb + (ub - lb) / 2; if (A[mid] < target) { lb = mid; } else { ub = mid; } } // whether A[lb + 1] == target, check lb + 1 first if ((lb + 1 < A.length) && (A[lb + 1] == target)) { result[0] = lb + 1; } else { result[0] = -1; result[1] = -1; // target is not in the array return result; } // upper bound, since ub >= lb, we do not reset lb ub = A.length; while (lb + 1 < ub) { int mid = lb + (ub - lb) / 2; if (A[mid] > target) { ub = mid; } else { lb = mid; } } // target must exist in the array result[1] = ub - 1; return result; } } ~~~ ### 源码分析 1. 首先对输入做异常处理，数组为空或者长度为0 1. 分 lower/upper bound 两次搜索，注意如果在 lower bound 阶段未找到目标值时，upper bound 也一定找不到。 1. 取`A[lb + 1]` 时一定要注意判断索引是否越界！ ### 复杂度分析 两次二分搜索，时间复杂度仍为 O(logn)O(\log n)O(logn).