边工作边刷题：70天一遍leetcode: day 24-1

Search a 2D Matrix

考点：方法很简单，对第0列和之后找到的行做两次2分，重点是用第一次二分后low和high的位置来确定行。对于二分没找到的情况：low这时候比high大1，low对应的是>target的元素，high对应的是<target的元素。对于这题，我们需要用high，因为行也是递增的。

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        m = len(matrix)
        n = len(matrix[0])
        low,high = 0,m-1
        while low<=high:
            mid = low + (high-low)/2
            if matrix[mid][0]==target:
                return True
            elif matrix[mid][0]>target:
                high=mid-1
            else:
                low=mid+1
        print high
        if high>=0 and high<m:
            row = high
            low,high = 0,n-1
            while low<=high:
                mid = low+(high-low)/2
                print mid
                if matrix[row][mid]==target:
                    return True
                elif matrix[row][mid]>target:
                    high=mid-1
                else:
                    low=mid+1
                
        return False