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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.
Given matrix
[
[1 ,5 ,7],
[3 ,7 ,-8],
[4 ,-8 ,9],
]
return [(1,1), (2,2)]
O(n3) time.
Thoughts:
If the matrix is Nx1, we can solve it easily like sum of contiguous subsequense. If it‘s Nx2, we just need to repeat the same process 3 times -- the first column, the second column and sum of the two columns as an Nx1 array. That‘s applicable to any cases. The key point is to traverse every possible combination of two columns with two for loops and calculate the sum of each columns and store them into a hashmap.
Code:
public int[][] submatrixSum(int[][] matrix) {
// Write your code here
int[][] res = new int[2][2];
int m = matrix.length;
if(m==0) return res;
int n = matrix[0].length;
for(int i=0;i<n;i++){
int[] sum = new int[m];
for(int j=i;j<n;j++){
for(int k=0;k<m;k++)
sum[k]+=matrix[k][j]; //traverse every possible combination of indices of each column
int lastSum=0;
HashMap<Integer,Integer> map = new HashMap<>();
map.put(0,-1);
for(int v=0;v<m;v++){
lastSum+=sum[v];
if(map.containsKey(lastSum)){
res[0][0]=map.get(lastSum)+1;
res[0][1]=i;
res[1][0]=v;
res[1][1]=j;
return res;
}
map.put(lastSum,v);
}
}
}
return res;
}
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原文地址：http://www.cnblogs.com/midan/p/4681715.html
