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LintCode Sliding Window Matrix Maximum
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原题链接在这里：http://www.lintcode.com/zh-cn/problem/sliding-window-matrix-maximum/
题目：
Given an array of
n * m
matrix, and a moving matrix window (size
k * k
), move the window from top left to botton right at each iteration, find the maximum number inside the window at each moving.
Return
0
if the answer does not exist.
For matrix
[
[1, 5, 3],
[3, 2, 1],
[4, 1, 9],
]
The moving window size k =
2
.
return 13.
At first the window is at the start of the array like this
[
[|1, 5|, 3],
[|3, 2|, 1],
[4, 1, 9],
]
,get the sum
11
;
then the window move one step forward.
[
[1, |5, 3|],
[3, |2, 1|],
[4, 1, 9],
]
,get the sum
11
;
then the window move one step forward again.
[
[1, 5, 3],
[|3, 2|, 1],
[|4, 1|, 9],
]
,get the sum
10
;
then the window move one step forward again.
[
[1, 5, 3],
[3, |2, 1|],
[4, |1, 9|],
]
,get the sum
13
;
SO finally, get the maximum from all the sum which is
13
.
题解：
用sum matrix来表示以[i,j]为右下角点时整个左上方的matrix的和. sum[i][j] = matrix[i-1][j-1] + sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1].
需要减掉sum[i-1][j-1]因为加重复了.
然后每次算以[i][j]为右下角, 边长为k的小matrix的和, sum[i][j] - sum[i-k][j] - sum[i][j-k] + sum[i-k][j-k].
需要加上sum[i-k][j-k]因为减重复了.
Time Complexity: O(m*n), m = matrix.length, n = matrix[0].length.
Space: O(m*n).
AC Java:
public class Solution {
public int maxSlidingMatrix(int[][] matrix, int k) {
if(matrix == null || matrix.length < k || matrix[0].length < k){
return 0;
}
int m = matrix.length;
int n = matrix[0].length;
int [][] sum = new int[m+1][n+1];
for(int i = 1; i<=m; i++){
for(int j = 1; j<=n; j++){
sum[i][j] = matrix[i-1][j-1] + sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];
}
}
int res = Integer.MIN_VALUE;
for(int i = k; i<=m; i++){
for(int j = k; j<=n; j++){
res = Math.max(res, sum[i][j]-sum[i-k][j]-sum[i][j-k]+sum[i-k][j-k]);
}
}
return res;
}
}
