LeetCode刷题【Array】Search a 2D Matrix

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本文介绍了一种高效的矩阵搜索算法，用于在一特殊性质的二维矩阵中查找特定值。该矩阵的每一行从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章提供了两种实现方式，一种是采用行内二分查找的方法，另一种是将矩阵视为一维数组进行搜索。

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 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.

解决方法一： Runtime: 1 ms 二分查找思想

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(null==matrix||matrix.length<=0||matrix[0].length<=0) return false;
		int start=0;
		int end=matrix.length-1;
		int row=0; int column=0;
		while(start<=end){
			row=(start+end)/2;
			if(matrix[row][column]==target) return true;
			else if(matrix[row][column]>target) end=row-1;
			else if(matrix[row][column]<target&&matrix[row][matrix[row].length-1]<target) start=row+1;
			else break;
		}
		start=0;end=matrix[row].length-1;
		while(start<=end){
			column=(start+end)/2;
			if(matrix[row][column]==target) return true;
			if(matrix[row][column]>target) end=column-1;
			if(matrix[row][column]<target) start=column+1;
		}
		return false;
    }
}

解决方法二：把二维数组当做Array处理，可能存在m*n Runtime: 1 ms

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

参考：

【1】https://leetcode.com/