LEETCODE: Search for a Range

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]

.

二分法查找的一个变种。

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> result;
        int left = 0;
        int right = n - 1;
        
        while(left <= right) {
            int mid = (left + right) >> 1;
            int midval = A[mid];
            
            if(midval == target) {
                int start = mid;
                while(start >= 0 && A[start] == midval) start --;
                start += 1;
                int end = mid;
                while(end < n && A[end] == midval) end ++;
                end -= 1;
                result.push_back(start);
                result.push_back(end);
                return result;
            }
            
            if(midval < target) {
                left = mid + 1;
            }
            else {
                right = mid - 1;
            }
        }
        
        result.push_back(-1);
        result.push_back(-1);
        return result;
    }
};