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]

题解

lower/upper bound 的结合，做两次搜索即可。

JAVA：

import java.io.*;
class test  
{
    public static void main (String[] args) throws java.lang.Exception
    {
        int arr[] = {5, 7, 7, 8, 8, 10};
        int target = 8;
        int range[] = searchRange(arr, target);
        System.out.println("range is " + range[0] + " - " + range[1]);
    }
    public static 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;
    }
}

输出：

range is 3 - 4

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

2. 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。

3. 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).