Submatrix Sum 405
Question
Given an integer matrix, find a submatrix where the sum of numbers is zero. Your code should return the coordinate of the left-up and right-down number.
Example
Given matrix
[ [1 ,5 ,7], [3 ,7 ,-8], [4 ,-8 ,9], ]
return [(1,1), (2,2)]
Challenge
O(n3) time.
Solution
这道题和求数组中哪些元素和为0的解决方法一样,只是数组中求的是前i个元素和前j个元素和相等,则i-j元素和为0,而这里只是变成2维的而已。
sum[i][j]表示matrix[0][0]到matrix[i-1][j-1]所有元素的和。
建立sum矩阵,为n+1行,m+1列。将第0行和第0列都初始化为0。
遍历matrix,根据公式 sum[i][j] = matrix[i - 1][j - 1] + sum[i][j - 1] + sum[i - 1][j] -sum[i - 1][j - 1] 计算所有sum。
然后取两个row:l1, l2。用一个线k从左到右扫过l1和l2,每次都用diff=sum[l1][k]-sum[l2][k]来表示l1-l2和0-k这个矩形元素的sum。如果在同一个l1和l2中,有两条线(k1,k2)的diff相等,则表示l1-l2和k1-k2这个矩形中的元素和为0。
代码如下:
public class Solution {
/**
* @param matrix an integer matrix
* @return the coordinate of the left-up and right-down number
*/
public int[][] submatrixSum(int[][] matrix) {
// Write your code here
int[][] res = new int[2][2];
if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
return res;
}
int m = matrix.length;
int n = matrix[0].length;
int[][] sum = new int[m + 1][n + 1];
for(int i = 0; i < m; i++){
sum[i][0] = 0;
}
for(int j = 0; j < n; j++){
sum[0][j] = 0;
}
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];
}
}
for(int l = 0; l < m; l++){
for(int h = l + 1; h <= m; h++){
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int k = 0; k <= n; k++){
int diff = sum[h][k] - sum[l][k];
if(map.containsKey(diff)){
////在matrix中和sum对应的点的index都要-1,在matrix中,左上角的点应该是循环中的点的右下方的点:l+1-1,map.get(diff)+1-1(即不包含循环中的左上角的点),而右下角的点是包含循环中右下角的点:h-1,k-1
res[0][0] = l;
res[0][1] = map.get(diff);
res[1][0] = h - 1;
res[1][1] = k - 1;
return res;
}else{
map.put(diff, k);
}
}
}
}
return res;
}
}