[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]

分析：排序数组，使用二分查找。

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int result_array[] = {-1, -1};
        vector<int> result(result_array, result_array + 2);
        if (nums.empty()) return result;
        
        int start = FindStart(nums, 0, nums.size() - 1, target);
        int end = FindEnd(nums, 0, nums.size() - 1, target);
        
        result[0] = start;
        result[1] = end;
        
        return result;
    }
    
    int FindStart(vector<int>& nums, int start, int end, int target) {
        if (start > end) return -1;
        
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) {
                end = mid;
            } else if (nums[mid] > target) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
        
        return nums[start] == target ? start : -1;
    }
    
    int FindEnd(vector<int>& nums, int start, int end, int target) {
        if (start > end) return -1;
        
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (start == end - 1) {
                if (nums[start] == target) {
                    if (nums[start] == nums[end])
                        return end;
                    else
                        return start;
                } 
            }
            if (nums[mid] == target) {
                start = mid;
            } else if (nums[mid] > target) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }
         return nums[end] == target ? end : -1;
    }
       
};