LeetCode-598. Range Addition II (JAVA)范围相加

598. Range Addition II

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 twopositive integers
a and b, which meansM[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:

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.

题目大意：

给定m * n矩阵M，初始为0，然后执行一些更新操作。

数组ops表示一组更新操作，每一个操作(a, b)，表示将矩阵0 <= i < a 并且 0 <= j < b的区域值+1。

进行若干操作后，求矩阵的最大值。

注意：

    m和n的范围[1, 40000]

    a的范围[1, m]，b的范围[1, n]

    操作不超过10000个

解题思路：

求ops[0 .. len][0]和ops[0 .. len][1]的最小值，

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。

public int maxCount(int m, int n, int[][] ops) {
if (ops == null || ops.length == 0) {
return m * n;
}

int row = Integer.MAX_VALUE;
int col = Integer.MAX_VALUE;
for (int[] op : ops) {
// op[0]是横坐标
row = Math.min(row, op[0]);
// op[1]是纵坐标
col = Math.min(col, op[1]);
}
// row * col是左上角的元素个数
return row * col;
}

public int maxCount(int m, int n, int[][] ops) {
// 初始化为m,n可以省略if，否则不可省
int row = m;
int col = n;
for (int[] op : ops) {
// op[0]是横坐标
row = Math.min(row, op[0]);
// op[1]是纵坐标
col = Math.min(col, op[1]);
}
// row * col是左上角的元素个数
return row * col;
}