#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; } };