leetcode------Search a 2D Matrix

标题：	Search a 2D Matrix
通过率	31.3%
难度	中等

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

本题思路清晰，就是再一个二维数组中找某个元素，那么立即想到了二分法，关键点，按行算，第k个元素的行数等于k/列数，列数等于k%列数。

还有一种思路就是，每一行的开头一定比上一行全部元素都大，那么从最后一行的第一个开始，每次比较每行的第一个，找到第一个比target小的元素，也就说明了target若存在一定在那一行，

1、二分法查找代码如下：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix==null || matrix.length==0 || matrix[0].length==0) 
            return false;
 
        int m = matrix.length;
        int n = matrix[0].length;
 
        int start = 0;
        int end = m*n-1;
 
        while(start<=end){
            int mid=(start+end)/2;
            int midX=mid/n;
            int midY=mid%n;
 
            if(matrix[midX][midY]==target) 
                return true;
 
            if(matrix[midX][midY]<target){
                start=mid+1;
            }else{
                end=mid-1;
            }
        }
 
        return false;
    }
}

2、查找第一个小的数：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m=matrix.length;
        int n=matrix[0].length;
        int i=m-1;
        int j=0;
        while(i>=0){
            if(matrix[i][0]>target)i--;
            else break;
        }
        if(i<0)return false;
        while(j<n){
            if(matrix[i][j]==target)return true;
            else j++;
        }
        return false;
        
    }
}

3、python代码：

class Solution:
    def searchMatrix(self, matrix, target):
        i, j = 0, len(matrix[0]) - 1
        while i < len(matrix) and j >= 0:
            if target < matrix[i][j]:
                j -= 1
            elif target > matrix[i][j]:
                i += 1
            else:
                return True
        return False