題目:

請看維基百科解釋
https://en.wikipedia.org/wiki/Pascal%27s_triangle
https://leetcode.com/problems/pascals-triangle/

解題思路:

利用2個for迴圈首個數字放入一，中間數值透過第二個for迴圈做相加在最後一個數字填入一，以此循環。

C版本:

int** generate(int numRows, int** columnSizes) {
    int i=0;    
    int j=0;
    if(numRows == 0)
        return 0;
    int ** returnArray = (int **)malloc(sizeof(int *) * numRows);
    *columnSizes = (int *)malloc(sizeof(int)*numRows);

    for(i=0; i<numRows; i++)
    {
        (*columnSizes)[i] = i+1;
         returnArray[i] = (int *)malloc(sizeof(int) * (i+1));
         for(j=0; j<i+1; j++)
         {
            if( (0 == j)  || (i == j) )
                returnArray[i][j] = 1;
            else
                returnArray[i][j] = returnArray[i-1][j-1] + returnArray[i-1][j];
         }
    }
    return returnArray;
}

Javascript版本:

var generate = function(n) {
  var cur = [1,1]
  var result = [1,1,1]
  var temp = []
  temp.push([1],[1,1]);
  if(n==1){return [1]}
  if(n==2){return result}

  for(var i = 2 ; i < n ; i ++){
    var next = [1]      
    for(var j = 0 ; j < cur.length-1  ; j ++){
        next.push(cur[j]+cur[j+1])
    }
    next.push(1)
    temp.push(next)
    cur = next;
  }
  return temp
};

時間複雜度為: O(nxn)

程式Github分享:

https://github.com/SIAOYUCHEN/leetcode

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本日分享:

One day, perhaps you will find the most touching thing is not that you have finished, but you finally have the courage to start.
有一天，或許你會發現， 最感動的不是你完成了， 而是你終於鼓起勇氣開始。