LeetCode-034. Search for a Range

1. 题目

Search for a Range

给定升序数组nums，找到给定target的起始和结束位置
算法的运行时复杂度必须是O(log n)
若找不到target，返回[-1, -1]

2. 分析

属于变形的二分查找中的其中两种情况
即查找第一个等于或者大于value的元素
和最后一个等于或者小于value的元素

因此可在找出l后，在[l,last)按顺序遍历或继续二分

3. 代码

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> result;
        result.push_back(-1);
        result.push_back(-1);

        int len = nums.size();
        if(len <= 0)
            return result;

        int l = searchFirstEqual(nums, target);
        if(l < 0 || l > nums.size())
            return result;

        int r = searchLastEqual(nums, l, target);
        if(r < 0 || r > nums.size())
            return result;

        // 不要写漏或者写反
        if(r < l)
            return result;

        result[0] = l;
        result[1] = r;
        return result;
    }

    int searchFirstEqual(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;

        while(left <= right)
        {
            int mid = left + ((right - left) >> 1);
            if(target > nums[mid])
                left = mid + 1;
            else
                right = mid - 1;
        }

        if(left >= 0 && left <= nums.size() - 1 && nums[left] == target)
            return left;

        return -1;
    }

    int searchLastEqual(vector<int>& nums, int l, int target) {
        int left = l;
        int right = nums.size() - 1;

        while(left <= right)
        {
            int mid = left + ((right - left) >> 1);
            if(target < nums[mid])
                right = mid - 1;
            else
                left = mid + 1;
        }

        if(right >= l && right <= nums.size() - 1 && nums[right] == target)
            return right;

        return -1;
    }
};

完整源代码放于github