leetcode:Search for a Range（数组，二分查找）

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：题意为在一个有序数组中找到给定目标值的起始位置并返回，如果目标值不存在则返回[1,1].

思路：使用binarySearchLow()去找到不小于目标值数字的最小索引，使用binarySearchUp()去找到不大于目标值数字的最大索引，然后即可得到索引范围。

code如下：

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class Solution { 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);     } 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;     } };

其他方法：先找到有序数组中与目标值相同的数字的位置，然后检查其个数.

code：

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class Solution { public:     vector<int> searchRange(vector<int>& nums, int target) {         int low=0;         int high=nums.size()-1;         vector<int> ans(2,-1);         int flag=-1;     //数组中是否存在与目标值相同数字的标志         int start,end;         while(low<=high){             int mid=(low+high)/2;             if(nums[mid]<target){                 low=mid+1;             }             else if(nums[mid]>target){                 high=mid-1;             }             else{                 flag=mid;                 break;             }         }         if(flag!=-1){             start=flag;             while(start>=0 && nums[start]==target){                 start--;             }             ans[0]=start+1;             end=flag;             while(end < nums.size() && nums[end]==target){                 end++;             }             ans[1]=end-1;         }         return ans;     } };

其他解法：解决问题的短代码（使用迭代器）

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class Solution { public:     vector<int> searchRange(vector<int>& nums, int target) {         vector<int> ret;         vector<int>::iterator start = find(nums.begin(), nums.end(), target);         vector<int>::reverse_iterator end = find(nums.rbegin(), nums.rend(), target);         ret.push_back( (start == nums.end() ? -1 : start-nums.begin() ) ),ret.push_back(nums.size() - 1 - (end - nums.rbegin()));         return ret;     } };

　　

 

　

python：

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class Solution(object):     def searchRange(self, nums, target):         """         :type nums: List[int]         :type target: int         :rtype: List[int]         """         if target not in nums:             return ([-1,-1])         low = nums.index(target)         nums.sort(reverse=True)         high = len(nums)-nums.index(target)-1         result = []         result.append(low)         result.append(high)         return (result)