[Leetcode] 34. Search for a Range 解题报告

题目：

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].

思路：

这还是典型的二分查找，关键在于怎么定义二分查找的边界条件。偷懒的做法是直接利用STL中提供的upper_bound和lower_bound方法，见代码片段1；不过估计这时签证官不会善罢甘休的，要求你写出bug free的upper_bound和lower_bound代码。那么请看代码片段2吧！

代码：

1、偷懒做法：

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> ret(2, -1);        auto it_start = lower_bound(nums.begin(), nums.end(), target);        auto it_end = upper_bound(nums.begin(), nums.end(), target);        if(it_start == it_end)            return ret;        ret[0] = distance(nums.begin(), it_start);        ret[1] = distance(nums.begin(), it_end) - 1;        return ret;    }};

2、完整做法：

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> ret = {-1, -1};        if(nums.size() == 0)            return ret;        int lower = lower_bound(nums, 0, nums.size() - 1, target);        int upper = upper_bound(nums, lower, nums.size() - 1, target);  // to accelerate        if(lower == upper)            return ret;        ret[0] = lower;        ret[1] = upper - 1;        return ret;    }private:    int lower_bound(vector<int>& nums, int left, int right, int target)    {        while(left <= right)        {            int mid = left + (right - left) / 2;            if(nums[mid] >= target)                right = mid - 1;            else                left = mid + 1;        }        return left;    }    int upper_bound(vector<int>& nums, int left, int right, int target)    {        while(left <= right)        {            int mid = left + (right - left) / 2;            if(nums[mid] > target)                right = mid - 1;            else                left = mid + 1;        }        return left;    }};