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34. Search for a Range

题目

Given an array of integers sorted in ascending order, 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]. 

解析

• 关于二分法的查找延伸的问题，细节要注意
• 注意使用stl的函数实现，在[first,last)区间查找

class Solution_34 {
public:
    int findUpBound(vector<int>& nums, int target) //找上边界
    {
        int low = 0, high = nums.size() - 1;
        if (nums.back()<target)
        {
            return -1;
        }
        while (low<=high)
        {
            int mid = (low + high) >> 1;
            if (nums[mid]<=target)   /// 向右夹逼，找到右边界
            {
                low = mid + 1;
            }
            else
            {
                high = mid - 1;
            }
        }
        return high; //向右夹逼，返回high
    }

    int findDownBound(vector<int>& nums, int target) // 找下边界
    {
        int low = 0, high = nums.size() - 1;
        if (nums.front()>target)
        {
            return -1;
        }
        while (low<=high)
        {
            int mid = (low + high) >> 1;
            if (nums[mid]<target)   /// 向左夹逼，则找到左边界
            {
                low = mid + 1;
            }
            else
            {
                high = mid - 1;
            }
        }

        return low; // 向左夹逼，返回low
    }

    vector<int> searchRange(vector<int>& nums, int target) { //找到target的上下边界
        vector<int> vec(2, -1);
        if (nums.size() == 0 || nums.front()>target || nums.back()<target)
        {
            return vec;
        }

        int high = findUpBound(nums, target);
        int low = findDownBound(nums, target);
        
        if ((low == high&&nums[low] == target) || (low<high))
        {
            vec[0] = low;
            vec[1] = high;
        }
        
        return vec;
    }
  

    vector<int> searchRange2(int A[], int n, int target) {
        vector<int > vec(A,A+n); //初始化

        return searchRange1(vec, target);

    }

    vector<int> searchRange1(vector<int>& nums, int target) { //找到target的自身上下边界，非严格上下界
        vector<int> vec(2, -1);
        if (nums.size() == 0||nums.front()>target||nums.back()<target)
        {
            return vec;
        }

        //pair<int, int> temp;
        //针对 upper_bound the objects in the range [first,last) are accessed. 本身传入的迭代器end().就是指向下一个位置
        auto range=equal_range(nums.begin(), nums.end(), target); //返回的是两个迭代器,解引用得到值
        
        if (*range.first!=target||*(range.second-1)!=target)
        {
            return vec; 
        }
        vec[0]=(range.first-nums.begin()); //stl: distance()计算迭代器之间的距离
        vec[1]=(range.second - nums.begin()-1);

        return vec;
    }
};

题目来源

• 34. Search for a Range