找LeetCode上简单的题目来撑过30天啦(DAY30)

题号:74 标题:Search a 2D Matrix 难度:Medium

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.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

Constraints:
• m == matrix.length
• n == matrix[i].length
• 1 <= m, n <= 100
• -104 <= matrix[i][j], target <= 104

我的程序码

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int i,j,k,lenx=matrix[0].length,leny=matrix.length;
        
        for(i=0;i<leny;){
            System.out.print("1" +i);
            if(target == matrix[i][lenx-1]){
                return true;
            }
            if(target > matrix[i][lenx-1]){
                System.out.print(" 2" +i);
                i++;
            }else{
                for(j=0;j<lenx-1;j++){
                    if(target == matrix[i][j]){
                        return true;
                    }
                }
                i++;
            } 
        }
        return false;
    }
}

day30心得
太好了，终於结束拉，终於不用每天都在想今天po了没，可喜可贺，也恭喜我30天每天都进步了一点点，今天的题目超级废XDD，题目并没有说要多有效率XDD我就近乎於暴力解啦，大家明年见罗