Today is an example of The non-Designer's Design Book on top of Page 56 and Reshape the Matrix on leetcode and leetcode's October LeetCoding Challenge with cpp.

The non-Designer's Design Book

my answer:

• [F] dashed font and dashed ellipse
• [F] same orange color
• [T] author and location use same font
• [F] same black color

Reshape the Matrix

Description

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.

Solution

basic cpp

class Solution {
public:
  vector<vector<int>> matrixReshape(vector<vector<int>>& nums, int r, int c) {
    if(nums.size() * nums[0].size() != r * c) return nums;
    vector<vector<int>> ans;
    vector<int> cur;
    int l = nums[0].size();
    for(int i = 0; i < r * c; ++i) {
      cur.push_back(nums[i/l][i%l]);
      if(cur.size() == c) {
        ans.push_back(cur);
        cur.clear();
      }
    }
    return ans;
  }
};

October LeetCoding Challenge 3

Description

K-diff Pairs in an Array Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

• 0 <= i, j < nums.length
• i != j
• a <= b
• b - a == k

Example 1:

Input: nums = [3,1,4,1,5], k = 2
Output: 2
Explanation: There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.

Example 2:

Input: nums = [1,2,3,4,5], k = 1
Output: 4
Explanation: There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).

Example 3:

Input: nums = [1,3,1,5,4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Example 4:

Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
Output: 2

Example 5:

Input: nums = [-1,-2,-3], k = 1
Output: 2

Constraints:

• 1 <= nums.length <= 104
• -107 <= nums[i] <= 107
• 0 <= k <= 107

Solution

simple two pointers

class Solution {
public:
  int findPairs(vector<int>& nums, int k) {
    sort(nums.begin(), nums.end());
    int ans = 0;
    if(k == 0) {
      int last_pair = -10000001;
      for (int i = 1; i < nums.size(); ++i) {
        if(nums[i] == nums[i-1] && nums[i] != last_pair) {
          ans += 1;
          last_pair = nums[i];
        }
      }
      return ans;
    }
    auto end = unique(nums.begin(), nums.end());
    auto front = nums.begin();
    for(auto back = nums.begin()+1; back != end; ++back) {
      while(front+1 != back && *back - *front > k) ++front;
      if(*back - *front == k) ++ans;
    }
    return ans;
  }
};