LeetCode 598. Range Addition II

You are given an m x n matrix M initialized with all 0's and an array of operations ops, where ops[i] = [ai, bi] means M[x][y] should be incremented by one for all 0 <= x < ai and 0 <= y < bi.

Count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:

Input: m = 3, n = 3, ops = [[2,2],[3,3]]
Output: 4
Explanation: The maximum integer in M is 2, and there are four of it in M. So return 4.

Example 2:

Input: m = 3, n = 3, ops = [[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3],[2,2],[3,3],[3,3],[3,3]]
Output: 4

Example 3:

Input: m = 3, n = 3, ops = []
Output: 9

Constraints:

• 1 <= m, n <= 4 * 104
• 0 <= ops.length <= 104
• ops[i].length == 2
• 1 <= ai <= m
• 1 <= bi <= n

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就，有点脑筋急转弯的感觉，主要难度在于理解题意……就是题目给出一系列ops[][]，每次要更新ops[i] = (a, b)范围内的矩阵的数字，求最后有多少个数字是每次都被更新了的。那么很显然每次只会更新左上角的数字，也就是求最小的ops[i][0]和ops[i][1]，然后把他们相乘就是最后结果。

Runtime: 3 ms, faster than 26.67% of Java online submissions for Range Addition II.

Memory Usage: 44.8 MB, less than 18.59% of Java online submissions for Range Addition II.

class Solution {
    public int maxCount(int m, int n, int[][] ops) {
        int row = m;
        int col = n;
        
        for (int[] op : ops) {
            row = Math.min(row, op[0]);
            col = Math.min(col, op[1]);
        }
        
        return row * col;
    }
}