[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.

思路：

要求统计矩阵最大值的个数，可以发现，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。

其实就是求a和b的最小值，然后框定的范围内的值一定都是最大的，个数就是n*m。

package com.billkang;/** * @author binkang * @date May 28, 2017 */public class RangeAddition2 {public int maxCount(int m, int n, int[][] ops) {int len = ops.length;if(len<1) {return m*n;}int min_a = Integer.MAX_VALUE;int min_b = Integer.MAX_VALUE;for(int i=0;i<len;i++) {int a = ops[i][0];int b = ops[i][1];min_a = Math.min(a, min_a);min_b = Math.min(b, min_b);}return min_a * min_b;}}