34.Search for a Range

思路：

• 二分查找，找到后分别向左向右遍历有多少重复的，然后记录左右端点即可,但是这样在某些情况下复杂度为(O(n))，比如{8,8,8,8,8,8,8,8,8}。竟然过了。。。

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int mid,lo = 0, hi = nums.size()-1;
        if(nums.size() == 0) return {-1,-1};
        int k = 0;
        std::vector<int> v;
        while(lo <= hi){
            mid = (lo + hi) / 2;
            cout << "ho " << hi << " lo " << lo << endl;
            cout << "mid " << mid << " nums[mid] " << nums[mid] << endl;
            if(nums[mid] == target){
                cout << target << endl;
                for(int i = mid-1; i >= 0; i--){
                    if(nums[i] == target) k++;
                }
                int left = mid - k;
                k = 0;
                for(int i = mid+1; i < nums.size(); i++){
                    if(nums[i] == target) k++;
                }
                int right = mid + k;
                v.push_back(left);
                v.push_back(right);
                return v;
            }
            else if(nums[mid] < target) lo = mid+1;
            else hi = mid-1;
        }
        
        return {-1,-1};
    }
};

• 用二分分别查找左端点和右端点，每次的复杂度都为(O(log(n)))。注意：1.要考虑是否找到；2.要考虑有没有越界(wa了好几次。。)

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int mid,lo = 0, hi = nums.size()-1;
         
        int k = 0;
        std::vector<int> v = {-1,-1};
        if(nums.size() == 0) return v;
        while(lo <= hi){
            mid = (lo + hi) / 2;
            if(nums[mid] < target)  lo = mid+1; 
            else    hi = mid-1;
        }
        int left = lo;
        if(left < nums.size() && nums[left] == target) v[0] = left;    //注意
        lo = 0,hi = nums.size()-1;
        while(lo <= hi){
            mid = (lo + hi) / 2;
            if(nums[mid] > target)  hi = mid-1;
            else    lo = mid + 1;
        }
        int right = hi;
        if(right >= 0 && nums[right] == target) v[1] = right;    //注意
        return v;
    }
};

总结：

1. 每次二分要判断是否找到
2. 测试程序要考虑到[],[a]两种特殊情况