2017/11/9 Leetcode 日记

Posted 2020-10-14 yoyo_sincerely

2017/11/9 Leetcode 日记

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.

(给一个矩阵和r, c，将这个矩阵重新排列成r行c列的矩阵，如果不可能则输出原矩阵。)

 

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

c++

class Solution:
    def matrixReshape(self, nums, r, c):
        """
        :type nums: List[List[int]]
        :type r: int
        :type c: int
        :rtype: List[List[int]]
        """
        row, col = len(nums), len(nums[0])
        if row * col != r * c:
            return nums
        o = r*c
        
        num = [[None] * c for _ in range(r)]
        for i in range(0, o):
            num[i//c][i%c] = nums[i//col][i%col]
        
        return num

python3

 

682. Baseball Game

You‘re now a baseball game point recorder.

Given a list of strings, each string can be one of the 4 following types:

1. Integer (one round‘s score): Directly represents the number of points you get in this round.
2. "+" (one round‘s score): Represents that the points you get in this round are the sum of the last two valid round‘s points.
3. "D" (one round‘s score): Represents that the points you get in this round are the doubled data of the last valid round‘s points.
4. "C" (an operation, which isn‘t a round‘s score): Represents the last valid round‘s points you get were invalid and should be removed.

 

Each round‘s operation is permanent and could have an impact on the round before and the round after.

You need to return the sum of the points you could get in all the rounds.

 

class Solution {
public:
    int calPoints(vector<string>& ops) {
        int len = ops.size();
        int sum = 0, index = 0;
        vector<int> op;
        
        for(int i = 0; i < len; i++){
            if (ops[i] == "+"){
                op.push_back(op[index-1] + op[index-2]);
            }else if (ops[i] == "C"){
                op.pop_back();
            }else if (ops[i] == "D"){
                op.push_back(2 * op[index-1]);
            }else{
                op.push_back(getNum(ops[i]));
            }
            index = op.size();
        }
        return getSum(op);
    }
    int getNum(string n){
        int num = 0;
        if(n[0] == ‘-‘){
            for(int i = 1, sz = n.size(); i<sz; i++){
                num *= 10;
                num += n[i]-‘0‘;
            }
            num = -num;
        }else{
            for(int i = 0, sz = n.size(); i<sz; i++){
                num *= 10;
                num += n[i]-‘0‘;
            }
        }
        return num;
    }
    int getSum(vector<int> op){
        int sum = 0; 
        for(int i = 0, sz = op.size(); i<sz; i++){
            sum += op[i];
        }
        return sum;
    }
};

c++

class Solution:
    def calPoints(self, ops):
        """
        :type ops: List[str]
        :rtype: int
        """
        OP = []
        index = 0
        for op in ops:
            if op == ‘+‘:
                OP.append(OP[index-1]+OP[index-2])
            elif op == ‘D‘:
                OP.append(2*OP[index-1])
            elif op == ‘C‘:
                OP.pop()
            else:
                OP.append(int(op))
        sum = 0
        for o in OP:
            sum += o
        return sum

Python3