Search for Range

题目：

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].

分析：

做两次binary search就好了。基本题。

解法：

class Solution {
    /** 
     *@param A : an integer sorted array
     *@param target :  an integer to be inserted
     *return : a list of length 2, [index1, index2]
     */
public:
    vector<int> searchRange(vector<int> &A, int target) {
        // write your code here
        vector<int> res = {-1, -1};
        if (A.size() == 0) {
            return res;
        } 

        int start = 0;
        int end = A.size() - 1;
        // find first
        while (start + 1 < end) {
            int 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) {
            res[0] = start;
        } else if (A[end] == target) {
            res[0] = end;
        } else {
            return res;
        }

        //find last
        start = 0;
        end = A.size() - 1;
        while (start + 1 < end) {
            int 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) {
            res[1] = end;
        } else if (A[start] == target) {
            res[1] = start;
        }

        return res;
    }
};