Pascal’s Triangle II

Description

Given an index k, return the k^th 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?

Tags: Array

思路0

题意是指定输出帕斯卡尔三角形的某一行，模拟即可，优化后的代码如下所示。

java:

class Solution {
    public List<Integer> getRow(int rowIndex) {
        List<Integer> res = new ArrayList<>();
        for (int i = 0; i <= rowIndex; ++i) {
            res.add(1);
            for (int j = i - 1; j > 0; --j) {
                res.set(j, res.get(j - 1) + res.get(j));
            }
        }
        return res;
    }
}

思路1

题目提示我们可以思考一下时间复杂度为O(n)的算法。仔细观察每一行数字即可找出规律。 kotlin(184ms/66.67%):

class Solution {
    fun getRow(rowIndex: Int): List<Int> {
        val res = Array(rowIndex + 1, { 1 })
        for (i in 1 until rowIndex + 1) {
            if (i > (rowIndex + 1) / 2) {
                res[i] = res[rowIndex - i]
                continue
            }
            res[i] = ((res[i - 1]) * (rowIndex - i + 1).toLong() / i).toInt()
        }
        return res.asList()
    }
}

JavaScript：

var getRow = function(rowIndex) {
  var rowArr = function(numRows) {
      let arr = []
      for(var i = 1; i <= numRows; i++) {
          if(i === 1) {
            arr.push([1])
            continue
          }
          if(i === 2) {
            arr.push([1, 1])
            continue
          }
          let innerArr = [1]
          for(var j = 2; j < i; j++) {
            innerArr.push(arr[i-2][j-2] + arr[i-2][j-1])
          }
          innerArr.push(1)
          arr.push(innerArr)
        }
        return arr
  }
  return rowArr(rowIndex + 1)[rowIndex]
};

结语

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