[LeetCode] 704. Binary Search

Description

Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1.

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1

Note:

1. You may assume that all elements in nums are unique.
2. n will be in the range [1, 10000].
3. The value of each element in nums will be in the range [-9999, 9999].

Analyse

从一个list里找出一个数，不存在则返回-1

使用二分查找，每次将问题的规模减半

int search(vector<int>& nums, int target)
{
    int left = 0;
    int right = nums.size() - 1;
    int mid = (left + right) / 2;

    while (left <= right)
    {
        if (nums[mid] == target)
        {
            return mid;
        }
        else if (nums[mid] > target)
        {
            right = mid - 1;
        }
        else if (nums[mid] < target)
        {
            left = mid + 1;
        }
        mid = (left + right) / 2;
    }

    return -1;
}

leetcode中最快的方法是调用STL中的lower_bound函数

template <class ForwardIterator, class T>
  ForwardIterator lower_bound (ForwardIterator first, ForwardIterator last, const T& val);

lower_bound在[first, last)的左闭右开区间寻找第一个不小于val的元素，采用了二分查找的思想

Returns an iterator pointing to the first element in the range [first,last) which does not compare less than val.

在这个区间找到满足条件的就返回指向这个值的iterator, 否则返回last，

An iterator to the lower bound of val in the range.
If all the element in the range compare less than val, the function returns last.

由于lower_bound是左闭右开的搜索，最后一个值未覆盖到，好在如果没找到回直接返回last，对lower_bound返回的值判断一下是否是target就可以覆盖对最后一个元素的搜索

int search(vector<int>& nums, int target)
{
        static int fast_io = []() { std::ios::sync_with_stdio(false); cin.tie(nullptr);
                                return 0; }();
    auto it = lower_bound(nums.begin(),nums.end(),target);
    return (it!=nums.end() && *it==target) ? it-nums.begin() : -1; //这里解决了val出现在最后的产生的问题
}

Reference

1. lower_bound - C++ Reference