leetcode 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 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 = 3operations = [[2,2],[3,3]]Output: 4Explanation: 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.

这道题本可以按照它的题意步骤一个个改数组元素，再统计最大值的个数，这当然是可以的。但是仔细看题意 0 <= i < a and 0 <= j < b 也就是说左上角的数是肯定会被覆盖到的，那么这么多operations一层一层地覆盖下来，我只要找到最小的覆盖面，这个部分每个operations都能覆盖到，那么这个部分的数字肯定是最大。

大神的想法也跟我是一样的。

package leetcode;public class Range_Addition_II_598 {public int maxCount(int m, int n, int[][] ops) {int minM=m;int minN=n;for(int i=0;i<ops.length;i++){int thisM=ops[i][0];int thisN=ops[i][1];minM=minM<=thisM?minM:thisM;minN=minN<=thisN?minN:thisN;}return minM*minN;}public static void main(String[] args) {// TODO Auto-generated method stubRange_Addition_II_598 r=new Range_Addition_II_598();int m=3;int n=3;int[][] ops=new int[][]{};System.out.println(r.maxCount(m, n, ops));}}

我原先代码中minM和minN的初始值都是Integer.MAX_VALUE，但是考虑到operations数组为空的情况的话需要直接返回m*n，那就让它们的初始值为m和n，这样所有情况都能照顾到。