Range Addition II
Range Addition II

Range Addition II
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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.```

Note:

1. The range of m and n is [1,40000].
2. The range of a is [1,m], and the range of b is [1,n].
3. The range of operations size won't exceed 10,000.

①只要找到二维坐标的第一个坐标的最小值和第二个坐标的最小值相乘即可；

public class Solution { //6ms
public int maxCount(int m, int n, int[][] ops) {
if(ops == null || ops.length == 0) return m * n;
int minRow = Integer.MAX_VALUE;
int minCol = Integer.MAX_VALUE;
for (int op[] : ops) {
if (minRow > op[0]) minRow = op[0];
if (minCol > op[1]) minCol = op[1];
}
return minRow * minCol;
}
}

② 我首先想到的是这种方法，按照题意依次将对应位置加1，然后遍历数组，查找与arr[0][0]相同的值的个数即可。但是出现了内存不足（Memory Limit Exceeded）。

public class Solution {
public int maxCount(int m, int n, int[][] ops) {
int[][] arr = new int[m][n];
for(int[] op : ops){
for(int i = 0;i < op[0];i ++){
for(int j = 0;j < op[1];j ++){
arr[i][j] += 1;
}
}
}
int count = 0;
for(int i = 0;i < m;i ++){
for(int j = 0;j < n;j ++){
if(arr[i][j] == arr[0][0]) count ++;
}
}
return count;
}
}

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