leetcode 119. Pascal's Triangle II
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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?
我思路是使用上一行来计算这一行的值。在同一个list中写入。
public List<Integer> getRow(int rowIndex) {List<Integer> list=new ArrayList<Integer>();if(rowIndex==0){list.add(1);return list;}list.add(1);list.add(1);for(int i=2;i<=rowIndex;i++){//第i行有i+1个数int tmp=1;for(int j=1;j<i;j++){int sum=tmp+list.get(j);tmp=list.get(j);list.set(j, sum);}list.add(1);}return list;}还有大神用了一种很妙的方法:一层层的杨辉三角,通过 arraylist 的 set 某个索引的值 来更新数组。
public List<Integer> getRow(int rowIndex) {List<Integer> list = new ArrayList<Integer>();if (rowIndex < 0)return list;for (int i = 0; i < rowIndex + 1; i++) {list.add(0, 1);for (int j = 1; j < list.size() - 1; j++) {list.set(j, list.get(j) + list.get(j + 1));}}return list;}
i=1 list: 1 1
i=2 list: 1 1 1 -> 1 2 1
i=3 list: 1 1 2 1 -> 1 3 3 1
i=4 list: 1 1 3 3 1 -> 1 4 6 4 1
i=5 list: 1 1 4 6 4 1 -> 1 5 10 10 5 1
是个很赞的方法!
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