34. Search for a Range (Array; Divide-and-Conquer)

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：先二分法找最左端，再二分法找最右端。保证稳定排序。具体实现：

1. 相同元素返回最左元素：start从-1开始，且总是在<target位置，最后会是最左侧=target元素之前的那个位置
2. 相同元素返回最右元素：end总是在>target位置，所以从n开始，最后会是最右侧=target元素之后的那个位置

结束条件：start+1==end

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        leftBinarySearch(nums,-1,nums.size()-1,target);//start始终在<target的位置
        if(result[0]!=-1) rightBinarySearch(nums,0,nums.size(),target);//end始终在>target的位置
        return result;
    }
    
    void leftBinarySearch(vector<int>& nums, int start, int end, int target){
        if(start+1==end){ //结束条件：只剩两个数（因为此时mid==start，会进入死循环）
            if(target == nums[end]){
                result.push_back(end);
            }
            else {
                result.push_back(-1);
                result.push_back(-1);
            }
            return;
        }
        
        int mid = start + ((end-start)>>1);
        if(target <= nums[mid]) leftBinarySearch(nums,start,mid,target); 
        else leftBinarySearch(nums,mid, end,target); //start始终在<target的位置
    }
    
    void rightBinarySearch(vector<int>& nums, int start, int end, int target){
        if(start+1==end){
            //must have one answer, so don't need if(target == nums[start])
            result.push_back(start);
            return;
        }
        
        int mid = start + ((end-start)>>1);
        if(target < nums[mid]) rightBinarySearch(nums,start,mid,target); //end始终在>target的位置
        else rightBinarySearch(nums,mid, end,target);
    }
private:
    vector<int> result;
};