61. 搜索区间

题目描述

给定一个包含 n 个整数的排序数组，找出给定目标值 target 的起始和结束位置。
如果目标值不在数组中，则返回[-1, -1]

样例
给出[5, 7, 7, 8, 8, 10]和目标值target=8,
返回[3, 4]

思路
直接用两次二分搜索，第一次寻找target出现的第一个位置，第二次寻找target出现的最后一个位置，时间复杂度为O(2logn)。如果直接搜索到一个target值在从该位置前后顺序搜索，最坏情况下耗时为O(nlogn),比如数组为{5，5，5，5，5，5.。。。。}

代码

public class Main {    /*    * @param A: an integer sorted array    * @param target: an integer to be inserted    * @return: a list of length 2, [index1, index2]    */    public static int[] searchRange(int[] A, int target) {        // write your code here        if (A.length == 0) {            return new int[]{-1, -1};        }        int start, end, mid;        int[] bound = new int[2];        // search for left bound        start = 0;        end = A.length - 1;        while (start +1 < end) {            mid = start + (end - start) / 2;            if (A[mid] == target) {                end = mid;            } else if (A[mid] < target) {                start = mid;            } else {                end = mid;            }        }        if (A[start] == target) {            bound[0] = start;        } else if (A[end] == target) {            bound[0] = end;        } else {            bound[0] = bound[1] = -1;            return bound;        }        // search for right bound        start = 0;        end = A.length - 1;        while (start +1 < end) {            mid = start + (end - start) / 2;            if (A[mid] == target) {                start = mid;            } else if (A[mid] < target) {                start = mid;            } else {                end = mid;            }        }        if (A[end] == target) {            bound[1] = end;        } else if (A[start] == target) {            bound[1] = start;        } else {            bound[0] = bound[1] = -1;            return bound;        }        return bound;    }}