边工作边刷题：70天一遍leetcode: day 35-4

Search for a Range

这两题的考点都是bounded binary search (i.e., > or >=), 程序结构要点：

• mid的更新在loop的最后，所以开始要初始化
• 条件是low<high，而不是low<=high, 否则可能infinite loop，因为low或者high可能不变，比如[1,3], 1：e.g., for >=: [4], mid0, low0, high0, target=3, 因为循环后low没有变化。同样这个例子说明为什么mid要在loop最后更新：lowhigh就跳出循环
• 对于low=mid, mid lean to high: mid = (low+high+1)/2

对于range这题，value不存在的corner cases还是比较复杂。首先，bounded binary search选择’>’ or ‘<‘，而不是’>=’ or ‘<=‘。这是因为如果>=，返回的左右边界不能用一个条件来判断是否是target的边界。另外，target超过左右边界元素的任意一个，都返回[-1,-1]。另一种情况是target就在边界上，这个corner case是在binary search里面处理，返回-1 or len(nums)，这样在主函数里就可以用同一个公式判断target是否存在。

class Solution(object):
    def searchRange(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        def binary_right(nums, target):
            if target>=nums[-1]: return len(nums)
            low,high=0,len(nums)-1
            mid = low+(high-low)/2
            while low<high:
                
                if nums[mid]>target:
                    high=mid
                else:
                    low=mid+1
                mid = low+(high-low)/2
            return mid
        
        def binary_left(nums, target):
            if target<=nums[0]: return -1
            low,high=0,len(nums)-1
            mid = (high+low+1)/2
            while low<high:
                if nums[mid]<target:
                    low=mid
                else:
                    high=mid-1
                mid = (high+low+1)/2
            return mid
            
        left,right = 0,0
        if not nums or nums[-1]<target or nums[0]>target:
            return [-1,-1]
            
        right=binary_right(nums, target)
        left=binary_left(nums, target)
        print "left=",left,"right=",right
        if nums[right-1]==target and nums[left+1]==target:
            return [left+1, right-1]
        return [-1,-1]