LeetCode 118/119. Pascal's Triangle 杨辉三角 (Java)

杨辉三角I

题目：

Given a non-negative integer numRows, generate the first numRows of Pascal’s triangle.

In Pascal’s triangle, each number is the sum of the two numbers directly above it.
Output:
[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]

解答：

class Solution {
    public List<List<Integer>> generate(int numRows) {
        List<List<Integer>> list=new ArrayList<>();
        if(numRows<=0){
            return list;
        }
        List<Integer> temp=new ArrayList<>();
        for(int i=1;i<=numRows;i++){
            temp.add(1);
            for(int j=temp.size()-2;j>0;j--){
                temp.set(j,temp.get(j-1)+temp.get(j));
            }
            list.add(new ArrayList<> (temp));   
        }
        return list;
    }
}

解析：

1. list.add(new ArrayList<> (temp))
   等于是复制了一个temp的副本出来，这个副本存在于list之中，里面的元素不会随着temp的改变而改变
   list.add(temp)
   加入的就是temp本身，如果temp改变了，那你从list获取这个temp的时候，也是得到改变的temp

杨辉三角II

题目：

Given a non-negative index k where k ≤ 33, return the kth index row of the Pascal’s triangle.

Note that the row index starts from 0.

In Pascal’s triangle, each number is the sum of the two numbers directly above it.

Example:
Input: 3
Output: [1,3,3,1]

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

解答：

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