#61 Search for a Range

题目描述：

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].

Have you met this question in a real interview? 

Yes

Example

Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Challenge 

O(log n) time.

题目思路：

这题我就用了简单的思路：找两次，第一次找最早出现target value的位置；第二次找最晚出现target value的位置。自觉code写的有些啰嗦，因为两段非常相似。。。

Mycode（AC = 49ms）：

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        int l = 0, r = A.size() - 1;        vector<int> ans(2, -1);                if (A.size() == 0) return ans;                // find the most left target value in A        while (l + 1 < r) {            int mid = (l + r) / 2;                        if (A[mid] < target) {                l = mid;            }            else if (A[mid] >= target) {                r = mid;            }        }                // searching from l to r        if (A[l] == target) {            ans[0] = l;        }        else if (A[r] == target) {            ans[0] = r;        }        else {            ans[0] = -1;        }                // find the most right target value in A        l = 0; r = A.size() - 1;        while (l + 1 < r) {            int mid = (l + r) / 2;                        if (A[mid] <= target) {                l = mid;            }            else if (A[mid] > target) {                r = mid;            }        }                // searching from r to l        if (A[r] == target) {            ans[1] = r;        }        else if (A[l] == target) {            ans[1] = l;        }        else {            ans[1] = -1;        }                return ans;    }};