LeetCode 240. Search a 2D Matrix II Java

题目：

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

题意：给出一个m*n的二维矩阵，具有以下特点：

　   (1)每一行按照升序排序

　　(2)每一列按照升序排列

现给出一个值，求该值是否在二维矩阵中存在。这里直接二次循环查找时间复杂度为o(m*n)，肯定会超时的。所以我先查找可能出现target的列和行，然后对可能的行和列分别做二分查找。

代码：

public class Solution {

	public boolean searchMatrix(int[][] matrix, int target) {
	
		if(matrix.length==0||matrix[0].length==0)		//空矩阵返回false
			return false;
		int row=matrix.length;
		int col=matrix[0].length;
		
		
		int mayRow=-1;		//可能的行
		int mayCol=-1;		//可能的列
		for(int i=0;i<row;i++){		//寻找可能出现target的行
			if(matrix[i][0]<=target&&matrix[i][col-1]>=target){
				mayRow=i;
				int low=0;
				int high=col-1;
				while(low<=high){
					int mid=(low+high)/2;
					if(matrix[mayRow][mid]==target){
						System.out.println("true");
						return true;
					}else if(matrix[mayRow][mid]>target){
						high=mid-1;
					}else if(matrix[mayRow][mid]<target){
						low=mid+1;
					}
				}
			}
		}
		
		for(int i=0;i<col;i++){	//寻找可能出现target的列
			if(matrix[0][i]<=target&&matrix[row-1][i]>=target){
				mayCol=i;
				int low=0;
				int high=row-1;
				while(low<=high){
					int mid=(low+high)/2;
					if(matrix[mid][mayCol]==target){
						System.out.println("true");
						return true;
					}else if(matrix[mid][mayCol]>target){
						high=mid-1;
					}else if(matrix[mid][mayCol]<target){
						low=mid+1;
					}
				}
			}
		}
		System.out.println("false");
		return false;
	}
}