5.17

Author: JiayangWu

0377.combination-sum-iv/combination-sum-iv.md

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+<p>Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.</p>
+
+<p><b>Example:</b></p>
+
+<pre>
+<i><b>nums</b></i> = [1, 2, 3]
+<i><b>target</b></i> = 4
+
+The possible combination ways are:
+(1, 1, 1, 1)
+(1, 1, 2)
+(1, 2, 1)
+(1, 3)
+(2, 1, 1)
+(2, 2)
+(3, 1)
+
+Note that different sequences are counted as different combinations.
+
+Therefore the output is <i><b>7</b></i>.
+</pre>
+
+<p>&nbsp;</p>
+
+<p><b>Follow up:</b><br />
+What if negative numbers are allowed in the given array?<br />
+How does it change the problem?<br />
+What limitation we need to add to the question to allow negative numbers?</p>
+
+<p><b>Credits:</b><br />
+Special thanks to <a href="https://leetcode.com/pbrother/">@pbrother</a> for adding this problem and creating all test cases.</p>

0377.combination-sum-iv/combination-sum-iv.py

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+class Solution(object):
+    def combinationSum4(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: int
+        """
+        dp = [0 for _ in range(target + 1)]
+        dp[0] = 1    
+        nums.sort() 
+        for i in range(1, target + 1):
+            for num in nums:
+                if i >= num:
+                    dp[i] += dp[i - num]
+                
+        return dp[target]

0423.reconstruct-original-digits-from-english/reconstruct-original-digits-from-english.md

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+<p>Given a <b>non-empty</b> string containing an out-of-order English representation of digits <code>0-9</code>, output the digits in ascending order.</p>
+
+<p><b>Note:</b><br />
+<ol>
+<li>Input contains only lowercase English letters.</li>
+<li>Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as "abc" or "zerone" are not permitted.</li>
+<li>Input length is less than 50,000.</li>
+</ol>
+</p>
+
+<p><b>Example 1:</b><br />
+<pre>
+Input: "owoztneoer"
+
+Output: "012"
+</pre>
+</p>
+
+<p><b>Example 2:</b><br />
+<pre>
+Input: "fviefuro"
+
+Output: "45"
+</pre>
+</p>

0423.reconstruct-original-digits-from-english/reconstruct-original-digits-from-english.py

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+class Solution(object):
+    def originalDigits(self, s):
+        """
+        :type s: str
+        :rtype: str
+        """
+        # zero one   two three  four    five   six seven       eight nine 
+        # z   o（先2  w  r(先4)  u      f(先4)  x  s（先6）    t(先3) 最后看e
+        #所以一个可行的查找的顺序是 two, four, six, one, three, five, seven, eight, nine
+        order = ["zero", "two", "four", "six", "one", "three", "five", "seven", "eight", "nine"]
+        find =  ["z", "w",    "u",   "x",    "o",  "r",     "f",    "v",     "t",     "e"]
+        digit = [0, 2, 4, 6, 1, 3, 5, 7, 8, 9]
+        
+        record = [0 for _ in range(10)]
+        dic = collections.Counter(s)
+        
+        for idx in range(10): #按digit数组的顺序遍历0~9
+            cnt = dic[find[idx]] #看看可以组成几个digit[idx]
+            record[digit[idx]] += cnt #记录下来
+            dic = dic - collections.Counter(order[idx] * cnt) #字典里减去对应的字母
+                
+            if not dic:
+                break
+            
+        ress = ""
+        for i in range(10): #转换成题目需要的格式
+            ress += str(i) * record[i]
+            
+        return ress
+        

0633.sum-of-square-numbers/sum-of-square-numbers.md

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+<p>Given a non-negative integer <code>c</code>, your task is to decide whether there&#39;re two integers <code>a</code> and <code>b</code> such that a<sup>2</sup> + b<sup>2</sup> = c.</p>
+
+<p><b>Example 1:</b></p>
+
+<pre>
+<b>Input:</b> 5
+<b>Output:</b> True
+<b>Explanation:</b> 1 * 1 + 2 * 2 = 5
+</pre>
+
+<p>&nbsp;</p>
+
+<p><b>Example 2:</b></p>
+
+<pre>
+<b>Input:</b> 3
+<b>Output:</b> False
+</pre>
+
+<p>&nbsp;</p>

0633.sum-of-square-numbers/sum-of-square-numbers.py

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+class Solution(object):
+    def judgeSquareSum(self, c):
+        """
+        :type c: int
+        :rtype: bool
+        """
+
+        if c == (int(c ** 0.5) ** 2):
+            return True
+        
+        lo, hi = 0, int(c ** 0.5)
+        while(lo <= hi):
+            s = lo ** 2 + hi ** 2
+            if s == c:
+                return True
+            elif s > c:
+                hi -= 1
+            else:
+                lo += 1
+        return False
+        

0912.sort-an-array/sort-an-array.md

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+<p>Given an array of integers <code>nums</code>, sort the array in ascending order.</p>
+
+<p>&nbsp;</p>
+
+<ol>
+</ol>
+
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong>&nbsp;[5,2,3,1]
+<strong>Output:</strong> [1,2,3,5]
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre>
+<strong>Input: </strong>[5,1,1,2,0,0]
+<strong>Output: </strong>[0,0,1,1,2,5]
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>Note:</strong></p>
+
+<ol>
+	<li><code>1 &lt;= A.length &lt;= 10000</code></li>
+	<li><code>-50000 &lt;= A[i] &lt;= 50000</code></li>
+</ol>

0912.sort-an-array/sort-an-array.py

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+class Solution(object):
+    def sortArray(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[int]
+        """
+        return sorted(nums)