Pascal's triangle I_II easy

Author: weiy

Diff for: Array/Pascal's TriangleI_II.py

@@ -0,0 +1,75 @@
+"""
+Given a non-negative integer numRows, generate the first numRows of Pascal's triangle.
+
+
+In Pascal's triangle, each number is the sum of the two numbers directly above it.
+
+Example:
+
+Input: 5
+Output:
+[
+     [1],
+    [1,1],
+   [1,2,1],
+  [1,3,3,1],
+ [1,4,6,4,1]
+]
+
+帕斯卡三角。
+
+图看链接：
+https://leetcode.com/problems/pascals-triangle/description/
+
+顺着思路来即可。
+
+I
+测试地址：
+https://leetcode.com/problems/pascals-triangle/description/
+
+II
+测试地址：
+https://leetcode.com/problems/pascals-triangle-ii/description/
+
+II 中的要求是返回最后一个，且空间为 O(K)。下面这个空间不是 O(k) 不过只要修改一下即可。
+
+"""
+# I
+class Solution(object):
+    def generate(self, numRows):
+        """
+        :type numRows: int
+        :rtype: List[List[int]]
+        """
+        result = []
+        
+        for i in range(1, numRows+1):
+            x = [0 for j in range(i)]
+            x[0] = 1
+            x[-1] = 1
+            for j in range(1, i-1):
+                x[j] = result[-1][j] + result[-1][j-1]
+            
+            result.append(x)
+        
+        return result
+
+# II
+ class Solution(object):
+    def generate(self, numRows):
+        """
+        :type numRows: int
+        :rtype: List[List[int]]
+        """
+        result = []
+        
+        for i in range(1, numRows+2):
+            x = [0 for j in range(i)]
+            x[0] = 1
+            x[-1] = 1
+            for j in range(1, i-1):
+                x[j] = result[-1][j] + result[-1][j-1]
+            
+            result.append(x)
+        
+        return result[-1]