Update

Author: sindwerra

leetcode/Greedy/#455-Assign Cookies/main.py

@@ -0,0 +1,61 @@
+# coding=utf-8
+
+'''
+Assume you are an awesome parent and want to give your children some cookies. 
+But, you should give each child at most one cookie. Each child i has a greed 
+factor gi, which is the minimum size of a cookie that the child will be content with; 
+and each cookie j has a size sj. If sj >= gi, we can assign the cookie j to the 
+child i, and the child i will be content. Your goal is to maximize the number of 
+your content children and output the maximum number.
+
+Note:
+You may assume the greed factor is always positive. 
+You cannot assign more than one cookie to one child.
+
+Example 1:
+Input: [1,2,3], [1,1]
+
+Output: 1
+
+Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 
+1, 2, 3. 
+And even though you have 2 cookies, since their size is both 1, you could only 
+make the child whose greed factor is 1 content.
+You need to output 1.
+Example 2:
+Input: [1,2], [1,2,3]
+
+Output: 2
+
+Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
+You have 3 cookies and their sizes are big enough to gratify all of the children, 
+You need to output 2.
+'''
+
+'''
+此题主要是证明正确性
+Beat 57.31%
+'''
+
+class Solution(object):
+    def findContentChildren(self, g, s):
+        """
+        :type g: List[int]
+        :type s: List[int]
+        :rtype: int
+        """
+        g.sort()
+        s.sort()
+        g_cursor = 0
+        s_cursor = 0
+        result = 0
+
+        while g_cursor < len(g) and s_cursor < len(s):
+            if g[g_cursor] > s[s_cursor]:
+                s_cursor += 1
+            else:
+                g_cursor += 1
+                s_cursor += 1
+                result += 1
+        
+        return result

leetcode/Matrix/#498-Diagonal Traverse/main.py

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+# coding=utf-8
+
+'''
+Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image.
+
+Example:
+Input:
+[
+ [ 1, 2, 3 ],
+ [ 4, 5, 6 ],
+ [ 7, 8, 9 ]
+]
+Output:  [1,2,4,7,5,3,6,8,9]
+Explanation:
+
+'''
+
+'''
+此题撞墙情况特别多，奇数轮和偶数轮坐标的移位都是有两种情况的，另外对角线打印的次数适合行列数总和有关的
+Beat 87.91%
+公司：Google
+'''
+
+class Solution(object):
+    def findDiagonalOrder(self, matrix):
+        """
+        :type matrix: List[List[int]]
+        :rtype: List[int]
+        """
+        n = len(matrix)
+        if not n:
+            return []
+
+        if n == 1:
+            return matrix[0]
+        
+        m = len(matrix[0])
+        if not m:
+            return []
+
+        if m == 1:
+            return [x[0] for x in matrix]
+
+        result = []
+        cursor = [0, 0]
+
+        for time in xrange(n + m - 1):
+            if time % 2 == 1:
+                while cursor[0] + 1 < n and cursor[1] - 1 >= 0:
+                    result.append(matrix[cursor[0]][cursor[1]])
+                    cursor[0] += 1
+                    cursor[1] -= 1
+                result.append(matrix[cursor[0]][cursor[1]])
+                if cursor[0] + 1 < n:
+                    cursor[0] += 1
+                else:
+                    cursor[1] += 1
+            else:
+                while cursor[0] - 1 >= 0 and cursor[1] + 1 < m:
+                    result.append(matrix[cursor[0]][cursor[1]])
+                    cursor[0] -= 1
+                    cursor[1] += 1
+                result.append(matrix[cursor[0]][cursor[1]])
+                if cursor[1] + 1 < m:
+                    cursor[1] += 1
+                else:
+                    cursor[0] += 1
+
+        return result