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;
}
}