Description:
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?
My code:
/**
* @param {number} rowIndex
* @return {number[]}
*/
var getRow = function(rowIndex) {
//第 n 行的第 k 个数字为组合数 C (k - 1)/(n - 1)
var getFactorial = function(num) { // 求阶乘
let factorial = 1;
for(let i = 1; i <= num; i++) {
factorial *= i;
}
return factorial;
};
var arr = [];
for(let i = 0; i <= rowIndex; i++) {
if(i == 0 || i == rowIndex) {
arr.push(1);
} else {
arr.push(getFactorial(rowIndex) / (getFactorial(i) * getFactorial(rowIndex - i)));
}
}
return arr;
};
Note: 题目说有没有办法优化到空间复杂度为O(k)，暂时还没想到……
