[LeetCode92]Search a 2D matrix

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本文介绍了一种高效的搜索算法，用于在一特殊性质的二维矩阵中查找指定值。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。文章详细解释了算法的工作原理，并提供了C++和Java两种实现方式。

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.

Analysis:

According to the rule of the matrix,

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

The minimum number is on the left top, the maximum number is on the right bottom.

so we begin search on the right top, if target>matrix[i][j] move down, else move left.

This algorithm taks O(m+n)

c++

bool searchMatrix(vector<vector<int> > &matrix, int target) {
    int row = matrix.size();
    if(row<=0) return false;
    int col = matrix[0].size();
    if(col<=0) return false;
    if(target<matrix[0][0] || target> matrix[row-1][col-1]) return false;
    int x1 = 0;
    int y1 = col-1;
    while(x1<=row-1 && y1>=0){
        if(target==matrix[x1][y1])
            return true;
        else if(target>matrix[x1][y1]){
            x1++;
        }else{
            y1--;
        }
    }
    return false;
}

java

public boolean searchMatrix(int[][] matrix, int target) {
		int row = matrix.length;
	    if(row<=0) return false;
	    int col = matrix[0].length;
	    if(col<=0) return false;
	    if(target<matrix[0][0] || target> matrix[row-1][col-1]) return false;
	    int x1 = 0;
	    int y1 = col-1;
	    while(x1<=row-1 && y1>=0){
	        if(target==matrix[x1][y1])
	            return true;
	        else if(target>matrix[x1][y1]){
	            x1++;
	        }else{
	            y1--;
	        }
	    }
	    return false;
	}