[LeetCode]118. Pascal's Triangle&119. Pascal's Triangle II

118 . Pascal’s Triangle

Easy

Given numRows, generate the first numRows of Pascal’s triangle.

For example, given numRows = 5,

Return

[

[1],

[1,1],

[1,2,1],

[1,3,3,1],

[1,4,6,4,1]

]

1ms:

public List<List<Integer>> generate(int numRows) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(numRows==0) return result;
List<Integer> first = new ArrayList<Integer>();
first.add(1);
result.add(first);
if(numRows==1) return result;
List<Integer> second = new ArrayList<Integer>();
second.add(1);second.add(1);
result.add(second);
if(numRows==2) return result;
for(int i=2;i<numRows;i++){
List<Integer> level = new ArrayList<Integer>();
level.add(1);
for(int j=1;j<i;j++){
level.add(result.get(i-1).get(j-1)+result.get(i-1).get(j));
}
level.add(1);
result.add(level);
}
return result;
}

119 . Pascal’s Triangle II

Easy

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?

2ms：

public List<Integer> getRow(int rowIndex) {
List<Integer> res = new ArrayList<>();
if (rowIndex < 0) return res;
// O(k) Memory
int[] prevLevel = new int[rowIndex+1];
for (int i = 0; i < rowIndex + 1; ++i) {
int diff = 0;
for (int index = 0; index <= i; ++index) {
if (index == 0 || index == i) {
prevLevel[i] = 1;
} else {
int newVal = prevLevel[index] + prevLevel[index-1] - diff;
diff = newVal - prevLevel[index];
prevLevel[index] =  newVal;
}
}
}

// Construct the result
for (int num : prevLevel) {
res.add(num);
}
return res;
}

Based on rules:

row k of Pascal’s Triangle:

[C(k,0), C(k,1), …, C(k, k-1), C(k, k)]

and

C[k,i] = C[k,i-1]*(k-i+1)/i

1ms:

public class Solution {
public List<Integer> getRow(int rowIndex) {
Integer[] rowList = new Integer[rowIndex+1];
rowList[0] = 1;
for(int i=1; i<rowList.length;i++) {
rowList[i] = (int)((long)rowList[i-1]*(rowIndex-(i-1))/(i));
}
return Arrays.asList(rowList);
}
}