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Search for a Range

Source

Problem

Given a sorted array of n integers, find the starting and ending position of a 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
  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)O(\log n).