Leetcode 0240: 搜索二维矩阵 II Search a 2D Matrix II

题目描述：

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The 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.

中文描述：

编写一个高效的算法来搜索 m x n 矩阵 matrix 中的一个目标值 target 。该矩阵具有以下特性：
每行的元素从左到右升序排列。
每列的元素从上到下升序排列。

Example 1:

Input: 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]], target = 5
Output: true

Example 2:

Input: 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]], target = 20
Output: false

Constraints:

m == matrix.length
n == matrix[i].length
1 <= n, m <= 300
-10⁹ <= matix[i][j] <= 10⁹
All the integers in each row are sorted in ascending order.
All the integers in each column are sorted in ascending order.
-10⁹ <= target <= 10⁹

Time complexity: O(n+m)
因为矩阵是从左往右，从上往下递增的，以左下交的点为起始点，则如果target的值大于当前的点就向右移动，target的值小于当前的点就向上移动。

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