[Leetcode] Search for a Range

题目：

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]

.

思路：要求复杂度为O(log n)，只能是二分法。具体方法是先用一次二分法找到target所在位置，然后在这个位置开始向左向右再做两次二分，找到边界。

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        int lower_bound = 0;
        int upper_bound = n - 1;
        vector<int> result(2, -1);
        int index = -1;
        while (lower_bound <= upper_bound) {
            int mid = (lower_bound + upper_bound) / 2;
            if (A[mid] == target) {
                index = mid;
                break;
            } else if (A[mid] < target) {
                lower_bound = mid + 1;
            } else {
                upper_bound = mid - 1;
            }
        }
        if (index < 0) return result;
        lower_bound = 0;
        upper_bound = index;
        int left_index = index;
        while (lower_bound <= upper_bound) {
            int mid = (lower_bound + upper_bound) / 2;
            if ((mid == 0 && A[mid] == target) || 
                (mid > 0 && A[mid] == target && A[mid-1] != target)) {
                left_index = mid;
                break;
            } else if (A[mid] == target) {
                upper_bound = mid - 1;
            } else {
                lower_bound = mid + 1;
            }
        }
        lower_bound = index;
        upper_bound = n - 1;
        int right_index = index;
        while (lower_bound <= upper_bound) {
            int mid = (lower_bound + upper_bound) / 2;
            if ((mid == n - 1 && A[mid] == target) || 
                (mid < n - 1 && A[mid] == target && A[mid+1] != target)) {
                right_index = mid;
                break;
            } else if (A[mid] == target) {
                lower_bound = mid + 1;
            } else {
                upper_bound = mid - 1;
            }
        }
        result[0] = left_index;
        result[1] = right_index;
        return result;
    }
};

总结：复杂度为O(log n).