leetcode——34——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].

自己写的代码，没有真正的实现二分查找，运行速度相对较慢

class Solution {public:vector<int> searchRange(vector<int>& nums, int target) {int i = 0;int j = nums.size() - 1;int mid;vector<int> res;while (i <= j){mid = i + (j - i) / 2;if (nums[mid] > target){j = mid - 1;}else if (nums[mid] < target){i = mid + 1;}else{ int m = mid, n = mid;while (nums[m] == target&&nums[m+1] ==target&&m<nums.size()-1){m++;}while (nums[n] == target&&nums[n-1] == target&&n>0){n--;}res.push_back(n);res.push_back(m);return res;}}res.push_back(-1);res.push_back(-1);return res;}};

完全实现二分查找的代码

class Solution {public:vector<int> searchRange(vector<int>& nums, int target) {vector<int> res(2, -1);        if(nums.empty()) return res;        int high = binarySearchUp(nums, target, 0, nums.size() -1);        int low = binarySearchLow(nums, target, 0, nums.size() - 1);        if(high >= low)        {            res[0] = low;            res[1] = high;            return res;        }        return res;}private:    int binarySearchLow(vector<int>& nums, int target, int begin, int end)    {        if(begin > end) return begin;        int mid = begin + (end - begin) / 2;        if(nums[mid] < target) return binarySearchLow(nums, target, mid + 1, end);        else return binarySearchLow(nums, target, begin, mid - 1);    }    int binarySearchUp(vector<int>& nums, int target, int begin, int end)    {        if(begin > end) return end;        int mid = begin + (end - begin) / 2;        if(nums[mid] > target) return binarySearchUp(nums, target, begin, mid - 1);        else return binarySearchUp(nums, target, mid + 1, end);    }};