题目

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]

分析

二分搜索法，先二分搜索左边界，再二分搜索右边界，注意搜索时边界的确定
搜索左边界的时候，end = mid,保证最后留下的两个数要么都等于target要么第一个小于
搜索右边界的时候，start = mid，保证最后留下的两个数，要么一个大于，要么两个都等于

代码

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) {
        // 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;
    }
}