my understanding of (lower bound,upper bound) binary search, in C++, thanks to two post 分类： leetcode 2015-08-01 14:35 113人阅读 评论(0) 收藏

If you understand the comments below, never will you make mistakes with binary search!

thanks to A simple CPP solution with lower_bound

and C++ O(logn) Binary Search that handles duplicate, thanks to phu1ku ‘s answer on the second post.

http://en.cppreference.com/w/cpp/algorithm/upper_bound

Returns an iterator pointing to the first element in the range [first, last) that is greater than value.

http://en.cppreference.com/w/cpp/algorithm/lower_bound

Returns an iterator pointing to the first element in the range [first, last) that is not less than (i.e. greater or equal to) value.

If want to practice, code on your own, try https://leetcode.com/problems/search-insert-position/ , https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ , https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/ and https://leetcode.com/problems/search-a-2d-matrix-ii/ , and see the discusses.

int binarySearch(vector<int>& nums, int target) {
    /// return index of first one that comp(item,target)==true, or nums.size() if not found
    /// comp is greater or equal to for lower_bound
    /// comp is greater for upper_bound
    int first=0, last=nums.size(), mid;
    while (first<last) {
        mid=first+((last-first)>>1); // first<=mid, mid<last
        /// if comp(item,target)==false, advance first
        // if(nums[mid]<=target) // for upper_bound
        if (nums[mid]<target) // for lower_bound
        first=mid+1; // first always increases
        else /// else recede last
        last=mid; // last always decreases (even last-first==1)
    }
    return first;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。// p.s. If in any way improment can be achieved, better performance or whatever, it will be well-appreciated to let me know, thanks in advance.