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Given an m * n matrix M initialized with all 0's and several update operations.
Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.
You need to count and return the number of maximum integers in the matrix after performing all the operations.
Example 1:
Input:
m = 3, n = 3
operations = [[2,2],[3,3]]
Output: 4
Explanation:
Initially, M =
[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]
After performing [2,2], M =
[[1, 1, 0],
[1, 1, 0],
[0, 0, 0]]
After performing [3,3], M =
[[2, 2, 1],
[2, 2, 1],
[1, 1, 1]]
So the maximum integer in M is 2, and there are four of it in M. So return 4.
注意事项
The range of m and n is [1,40000].
The range of a is [1,m], and the range of b is [1,n].
The range of operations size won't exceed 10,000.
解题思路1：
我一开始什么都没想，直接按照题目的要求一步一步模仿写出过程代码，结果33%时内存溢出了。。。
public class Solution {
/**
* @param m: an integer
* @param n: an integer
* @param ops: List[List[int]]
* @return: return an integer
*/
public int maxCount(int m, int n, int[][] ops) {
// write your code here
int[][] matrix = new int[m][n];
int max = 0;//最大整数
//对矩阵按照题目要求初始化，并求出矩阵中的最大元素
for(int i=0 ; i<ops.length ; i++){
for(int j=0 ; j<ops[i].length ; j++){
for(int row=0 ; row<ops[i][0] ; row++){
for(int col=0 ; col<ops[i][1] ; col++){
matrix[row][col]++;
if(matrix[row][col] > max)
max = matrix[row][col];
}
}
}
}
int res = 0;
//获取最大元素的次数
for(int i=0 ; i<matrix.length ; i++){
for(int j=0 ; j<matrix[i].length ; j++){
if(max == matrix[i][j])
res++;
}
}
return res;
}
}解题思路2：
后来一想，矩阵越靠近左上角的元素值越大，因为要加1的元素 行和列索引是从0开始的。 那么只需要找到操作次数最多的元素位置即可。而操作次数最多的元素肯定是偏向于靠近矩阵左上角的。所以只需找到ops的最小行与最小列即可。
public class Solution {
/**
* @param m: an integer
* @param n: an integer
* @param ops: List[List[int]]
* @return: return an integer
*/
public int maxCount(int m, int n, int[][] ops) {
// write your code here
int row = m;
int col = n;
for(int[] op : ops){
row = Math.min(row,op[0]);// op[0]是横坐标
col = Math.min(col,op[1]);// op[1]是纵坐标
}
return row*col;
}
}