LeetCode(034) Search For a Range (Java)

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题目如下：

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].

分析如下：

1. 首先 while (start + 1 < end) 这个固定套路。

2. 我使用了

                if(A[mid - 1] < target)  {                    return mid;

来表达左边界(右边界同理)

3. 二分搜索结束后，判断return -1的情况。

我的代码：

// 242 mspublic class Solution {        public int searchStart(int[] A, int target) {        int start = 0, end = A.length - 1, result = -1, mid = 0;        while (start + 1 < end) {            mid = start + (end - start) / 2;            if(A[mid] > target) {                end = mid - 1;            } else if (A[mid] < target) {                start = mid + 1;            } else {                if(A[mid - 1] < target)  {                    return mid;                } else {                    end = mid;                }            }        }        if (A[start] == target) return start;        else if (A[end] == target) return end;        else return -1;    }        public int searchEnd(int[] A, int target) {        int start = 0, end = A.length - 1, result = -1, mid = 0;        while (start + 1 < end) {            mid = start + (end - start) / 2;            if(A[mid] > target) {                end = mid - 1;            } else if (A[mid] < target) {                start = mid + 1;            } else {                if(A[mid + 1] > target)  {                    return mid;                } else {                    start = mid;                }            }        }        if (A[end] == target) return end;        else if (A[start] == target) return start;        else return -1;            }        public int[] searchRange(int[] A, int target) {        int [] result = new int[2];        result[0] = searchStart(A, target);        result[1] = searchEnd(A, target);        return result;    }}

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