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LeetCode之Pascal's Triangle II

原題:

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

題意:返回揚輝三角第K行數,從0開始記


最笨的想法:

用兩個陣列,計算時用到上次陣列,程式碼如下:結果是通過的

public List<Integer> getRow(int rowIndex) {

		List<Integer> list = new ArrayList<>();

		if (rowIndex == 0) {
			list.add(new Integer(1));
			return list;
		} else if (rowIndex == 1) {
			list.add(new Integer(1));
			list.add(new Integer(1));
			return list;
		} else {

			List<Integer> oldList = new ArrayList<>(); //從第2行開始
			oldList.add(new Integer(1));
			oldList.add(new Integer(1));
			
			for (int i = 1; i <= rowIndex; i++) {
				list = new ArrayList<>();
				list.add(new Integer(1));			
				for (int j = 1; j < i ; j++) {
					list.add(oldList.get(j) + oldList.get(j -1));
				}
				list.add(new Integer(1));
				oldList = list;
			}			
			return list;
		}
	}
	


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