zoukankan      html  css  js  c++  java
  • leetcode 566. Reshape the Matrix

    In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new one with different size but keep its original data.

    You're given a matrix represented by a two-dimensional array, and two positive integers r and c representing the row number and column number of the wanted reshaped matrix, respectively.

    The reshaped matrix need to be filled with all the elements of the original matrix in the same row-traversing order as they were.

    If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.

    Example 1:

    Input: 
    nums = 
    [[1,2],
     [3,4]]
    r = 1, c = 4
    Output: 
    [[1,2,3,4]]
    Explanation:
    The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.

    Example 2:

    Input: 
    nums = 
    [[1,2],
     [3,4]]
    r = 2, c = 4
    Output: 
    [[1,2],
     [3,4]]
    Explanation:
    There is no way to reshape a 2 * 2 matrix to a 2 * 4 matrix. So output the original matrix.

    Note:

    1. The height and width of the given matrix is in range [1, 100].
    2. The given r and c are all positive.

    这个题目好无聊

    class Solution {
    public:
        vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
            vector<vector<int> > w;
            vector<int> v;
            if (nums.size() == 0 || (r * c > nums.size()*nums[0].size())) return nums;
            for (int i = 0; i < nums.size(); ++i) {
                for (int j = 0; j < nums[i].size(); ++j) {
                    v.push_back(nums[i][j]);
                    if (v.size() == c) {
                        w.push_back(v);
                        v.clear();
                    }
                }
            }
            return w;
        }
    };
  • 相关阅读:
    JAVA并发编程学习笔记之ReentrantLock
    服务架构演进
    Java集群优化——dubbo+zookeeper构建高可用分布式集群
    Dubbo实例
    hessian学习
    JAVA分布式事务原理及应用
    了解AngularJS $resource
    AngularJS Resource:与 RESTful API 交互
    Hibernate解决高并发问题之:悲观锁 VS 乐观锁
    互联网金融高并发方案
  • 原文地址:https://www.cnblogs.com/pk28/p/7248543.html
Copyright © 2011-2022 走看看