LeetCode——Search a 2D Matrix II

Description：

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 in ascending from left to right.

Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following 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]
]

Given target =

5

, return

true

.

Given target =

20

, return

false

.

首先想到的就是遍历整个矩阵，时间复杂度是O(m*n)，肯定是Timeout。

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {

for(int i=0; i<matrix.length; i++) {
for(int j=0; j<matrix[i].length; j++) {
if(matrix[i][j] == target) {
return true;
}
}
}

return false;
}
}

然后在优化的话就想到了二分。对每一行进行二分。时间复杂度是O(n*logm)，还是Timeout，二分用的越多时间复杂度就越高，所以行列都二分（有点递归分治的意思）更会Timeout。

public class Solution {

public boolean binarySearch(int[] arr, int terget) {

int left = 0, int right = arr.length - 1;
while(left <= right) {
int mid = left + (right - left) / 2;
if(target == arr[mid]) {
return true;
}
if(target > arr[mid]) {
left = mid + 1;
}
else {
right = mid - 1;
}
}

return false;
}

public boolean searchMatrix(int[][] matrix, int target) {

for(int i=0; i<matrix.length; i++) {
if(binarySearch(matrix[i], target)) {
return true;
}
}

return false;
}

}

这么看来时间复杂度必须在线性的基础上才行。观察一下矩阵不难发现把矩阵逆时针旋转45度类似一棵二叉查找树。所以就可以模仿二叉查找树的方法来做了。

这样的话时间复杂度就是O(m + n)；AC。

public class Solution {

public boolean searchMatrix(int[][] matrix, int target) {

if(matrix.length==0 || matrix[0].length==0) {
return false;
}

int i = 0, j = matrix[0].length - 1;

while(i < matrix.length && j >= 0) {
int cur = matrix[i][j];
if(cur == target) {
return true;
}
else if(cur < target) {
i ++;
}
else {
j --;
}

}
return false;

}

}