[LeetCode] 119. Pascal's Triangle II 杨辉三角 II

Given an index k, return the kth row of the Pascal's triangle.

For example, given k = 3,
Return [1,3,3,1].

Note:
Could you optimize your algorithm to use only O(k) extra space?

118. Pascal's Triangle 的拓展，给一个索引k，返回杨辉三角的第k行。

解法：题目要求优化到 O(k) 的空间复杂，那么就不能把每行都记录下来，而只是记录前一行和当前行。

Java:

public class Solution {  
    public List<Integer> getRow(int rowIndex) {  
        List<Integer> res = new ArrayList<>();  
        int curr[] = new int[rowIndex + 1];  
        int prev[] = new int[rowIndex + 1];  
        prev[0] = 1;  
      
        for(int row = 1; row <= rowIndex; row++) {  
            curr[0] = 1;  
            curr[row] = 1;  
            for(int i = 1; i < row; i++)  
                curr[i] = prev[i] + prev[i - 1];  
            int[] swap = curr;  
            curr = prev;  
            prev = swap;  
        }   
      
        for(int i = 0; i <= rowIndex; i++)  
            res.add(prev[i]);  
        return res;  
    }  
} 　

Java:

public class Solution {
    public List<Integer> getRow(int rowIndex) {
        ArrayList<Integer> row = new ArrayList<Integer>();
        for (int i=0; i<rowIndex+1; i++){
            row.add(0,1);
            for(int j=1; j<row.size()-1;j++){
                row.set(j, row.get(j)+row.get(j+1));
            }
        }
        return row;
    }
}　　

Python:

class Solution:
    # @return a list of integers
    def getRow(self, rowIndex):
        result = [0] * (rowIndex + 1)
        for i in xrange(rowIndex + 1):
            old = result[0] = 1
            for j in xrange(1, i + 1):
                old, result[j] = result[j], old + result[j]
        return result

C++:　　

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> out;
        if (rowIndex < 0) return out;
        
        out.assign(rowIndex + 1, 0);
        for (int i = 0; i <= rowIndex; ++i) {
            if ( i == 0) {
                out[0] = 1;
                continue;
            }
            for (int j = rowIndex; j >= 1; --j) {
                out[j] = out[j] + out[j-1];
            }
        }
        return out;
    }
};

C++:

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int> result(rowIndex+1, 0);
        result[0] = 1;
        for(int i=1; i<rowIndex+1; i++)
            for(int j=i; j>=1; j--)
                result[j] += result[j-1];
        return result;
    }
};　　

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[LeetCode] 118. Pascal's Triangle 杨辉三角

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