zoukankan      html  css  js  c++  java
  • 598. Range Addition II 范围相加

    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 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:

    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.
    题意:给定一个m*n的矩阵,把数组初始化为0,然后用ops二维数组去更新矩阵的每个元素值,例如a=2,b=2则矩阵M11-M22全部增加1,以此类推,然后找出最大的元素的个数。
    解法:寻找到被操作次数最多的行和列
    1. public class Solution {
    2. public int MaxCount(int m, int n, int[,] ops) {
    3. int rowMin = Int32.MaxValue;
    4. int colMin = Int32.MaxValue;
    5. if (ops.GetLength(0) == 0) {
    6. return m * n;
    7. }
    8. for (int i = 0; i < ops.GetLength(0); i++) {
    9. int row = ops[i, 0];
    10. int col = ops[i, 1];
    11. if (row > 0) {
    12. rowMin = Math.Min(rowMin, row);
    13. } else {
    14. return 0;
    15. }
    16. if (col > 0) {
    17. colMin = Math.Min(colMin, col);
    18. }else {
    19. return 0;
    20. }
    21. }
    22. return rowMin * colMin;
    23. }
    24. }







  • 相关阅读:
    jm8.6编解码器概述
    mingw32环境下链接库找不到问题
    ts流中的pcr与pts计算与逆运算
    基于医疗知识图谱的问答系统(二)
    Neo4j图数据库导入数据
    基于医疗知识图谱的问答系统(一)
    知识图谱和neo4j的基本操作
    从.NET转GO了
    Flask开发技巧之参数校验
    如何在PPT中插入Pyecharts的图表?
  • 原文地址:https://www.cnblogs.com/xiejunzhao/p/2ef28b4838e32b29f4a8886245304564.html
Copyright © 2011-2022 走看看