# leetcode-119-easy()-其他

## leetcode-119-easy()

Pascal’s Triangle II

``````Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal's triangle.

In Pascal's triangle, each number is the sum of the two numbers directly above it as shown:

Example 1:

Input: rowIndex = 3
Output: [1,3,3,1]
Example 2:

Input: rowIndex = 0
Output: [1]
Example 3:

Input: rowIndex = 1
Output: [1,1]
Constraints:

0 <= rowIndex <= 33
Follow up: Could you optimize your algorithm to use only O(rowIndex) extra space?
``````

``````public List<Integer> getRow(int rowIndex) {
int[] arr = {1};
int[] result = arr;

for (int i = 2; i <= rowIndex + 1; i++) {
result = new int[i];

result[0] = 1;
result[i - 1] = 1;
for (int j = 1; j < i - 1; j++) {
result[j] = arr[j] + arr[j - 1];
}
arr = result;
}

List<Integer> list = new ArrayList<>();
for (int r : result) {
}
return list;
}
``````
————————

Pascal’s Triangle II

``````Given an integer rowIndex, return the rowIndexth (0-indexed) row of the Pascal's triangle.

In Pascal's triangle, each number is the sum of the two numbers directly above it as shown:

Example 1:

Input: rowIndex = 3
Output: [1,3,3,1]
Example 2:

Input: rowIndex = 0
Output: [1]
Example 3:

Input: rowIndex = 1
Output: [1,1]
Constraints:

0 <= rowIndex <= 33
Follow up: Could you optimize your algorithm to use only O(rowIndex) extra space?
``````

``````public List<Integer> getRow(int rowIndex) {
int[] arr = {1};
int[] result = arr;

for (int i = 2; i <= rowIndex + 1; i++) {
result = new int[i];

result[0] = 1;
result[i - 1] = 1;
for (int j = 1; j < i - 1; j++) {
result[j] = arr[j] + arr[j - 1];
}
arr = result;
}

List<Integer> list = new ArrayList<>();
for (int r : result) {