LeetCode – Search a 2D Matrix (Java)

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本文介绍了一种基于二分查找的高效算法，用于在具有特定性质的二维矩阵中搜索目标值。矩阵特性包括每行从左到右递增排序，且每行的第一个元素大于上一行的最后一个元素。通过巧妙地将矩阵视为一个有序数组，实现快速定位目标值。

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

1) Integers in each row are sorted from left to right. 2) 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.

Java Solution

This is a typical problem of binary search.

You may try to solve this problem by finding the row first and then the column. There is no need to do that. Because of the matrix’s special features, the matrix can be considered as a sorted array. Your goal is to find one element in this sorted array by using binary search.

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;
    }
}