62 lines
1.3 KiB
Markdown
62 lines
1.3 KiB
Markdown
# 搜索二维矩阵 Ⅱ
|
||
|
||
## 题目
|
||
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。
|
||
|
||
|
||
示例 1：
|
||
|
||
|
||
输入：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
|
||
输出：true
|
||
示例 2：
|
||
|
||
|
||
输入：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
|
||
输出：false
|
||
|
||
|
||
提示：
|
||
|
||
m == matrix.length
|
||
n == matrix[i].length
|
||
1 <= n, m <= 300
|
||
-109 <= matrix[i][j] <= 109
|
||
每行的所有元素从左到右升序排列
|
||
每列的所有元素从上到下升序排列
|
||
-109 <= target <= 109
|
||
|
||
|
||
## 思路
|
||
|
||
- 起点
|
||
- 右上角
|
||
|
||
## 代码
|
||
|
||
```java
|
||
class Solution {
|
||
public boolean searchMatrix(int[][] matrix, int target) {
|
||
int m = matrix.length; // height
|
||
int n = matrix[0].length; // wide
|
||
// Traverse: (Init: Location)
|
||
int loc_i = 0;
|
||
int loc_j = n - 1;
|
||
|
||
while (loc_i < m && loc_j >= 0) {
|
||
if (matrix[loc_i][loc_j] == target) {
|
||
return true;
|
||
} else if (matrix[loc_i][loc_j] < target) {
|
||
loc_i++;
|
||
} else {
|
||
loc_j--;
|
||
}
|
||
}
|
||
|
||
return false;
|
||
}
|
||
}
|
||
``` |