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：

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：

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）

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：

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)