[LeetCode] Search for a Range

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].

 

使用两次二分查找，第一次找出目标数的左边界，第二次找出目标数的右边界。最后判断边界并返回。

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        if (nums.empty())
            return {-1, -1};
        int lower = -1, upper = -1, mid = 0;
        
        int left = 0, right = nums.size() - 1;
        while (left <= right) {
            mid = (left + right) / 2;
            if (nums[mid] < target)
                left = mid + 1;
            else
                right = mid - 1;
        }
        lower = left;
        
        left = 0, right = nums.size() - 1;
        while (left <= right) {
            mid = (left + right) / 2;
            if (nums[mid] > target) 
                right = mid - 1;
            else
                left = mid + 1;
        }
        upper = right;
        
        if (lower > upper)
            return {-1, -1};
        else 
            return {lower, upper};
    }
};
// 16 ms

 

使用STL中的函数来完成这个算法

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        // judge nums'empty
        if (nums.empty())
            return {-1, -1};
        // find iterator that its value is equal or greater than target.
        auto left = lower_bound(nums.begin(), nums.end(), target);
        // find iterator that its value is greater than target.
        auto right = upper_bound(nums.begin(), nums.end(), target);
        
        // if target is not in nums.
        if ((left == right && *left != target) || (left == nums.end())) {
            return {-1, -1};
        }
        else {
            // find the iterator that its value is equal to target.
            right = prev(right);
            // distance is to calc the index in nums.
            return {distance(nums.begin(), left), distance(nums.begin(), right)};
        }
    }
};
// 9 ms