Pascal's Triangle 2(leetcode java)
问题描述:
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?
问题分析:
第 rowIndex 行有 rowIndex + 1 个元素,给这 rowIndex + 1个元素赋初值 0,
row的第一个元素赋值为1,此时,row的内容为[1,0,0,0,......]
由第 j 行生成第j + 1行,初始为[1,0,0,0,......],逆序将相邻两个元素相加后,赋值给后者
j == 0 时,生成的是 [1,1,0,0,......]
j == 1 时,生成的是 [1,2,1,0,......]
......
j == rowIndex - 1,生成最终结果
代码:
public List<Integer> getRow(int rowIndex) {
List<Integer> row = new ArrayList<Integer>();
if(rowIndex < 0) return row;
// 第 rowIndex 行有 rowIndex + 1 个元素,给这 rowIndex + 1个元素赋初值 0
for(int i = 0; i <= rowIndex; ++i){
row.add(0);
} row.set(0, 1); //生成第rowIndex行数据,最小为第0行
//由第 j 行逆序生成第j + 1行,初始为 1
//j == 0 时,生成的是 1 1
//j == 1 时,生成的是 1 2 1
//......
//j == rowIndex - 1,生成最终结果
for(int j = 0; j < rowIndex; ++j){
for(int k = rowIndex - 1; k > 0; --k){ //从最后一个元素开始,逆序计算
row.set(k, row.get(k) + row.get(k - 1));
}
}
row.set(rowIndex, 1); //给最后一位赋值1
return row;
}
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