LeetCode Range Addition II Given an m * n matrix M initialized with all 0‘s and several update operations. Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b. You need to count and return the number of maximum integers in the matrix after performing all the operations. Example 1:

Input:
m = 3, n = 3
operations = [[2,2],[3,3]]
Output: 4
Explanation:
Initially, M =
[[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]]
After performing [2,2], M =
[[1, 1, 0],
 [1, 1, 0],
 [0, 0, 0]]
After performing [3,3], M =
[[2, 2, 1],
 [2, 2, 1],
 [1, 1, 1]]
So the maximum integer in M is 2, and there are four of it in M. So return 4.

Note:

1. The range of m and n is [1,40000].
2. The range of a is [1,m], and the range of b is [1,n].
3. The range of operations size won’t exceed 10,000.

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给定一个全为0的矩阵M，和一系列的操作，每个操作都是一对(a,b)，表示要对所有i在[0,a)和j在[0,b)的元素M[i][j]，问最终矩阵M中最大元素的个数。 这个题稍微想一下就会发现非常简单。 假设矩阵的左下角坐标为(0,0)，对于某一个操作(a,b)，表示把以(0,0)和(a,b)为对角顶点的子矩阵元素加1。那么(a1,b1)和(a2,b2)两个操作的重叠区域(0,0)-(a2,b1)所在子矩阵的元素就被加了两次，他们的值最大且相同。 所以我们只需要遍历所有操作，分别找到行和列的最小值，则他们和原点围成的子矩阵的元素值最大，且相等。 代码非常简单，如下： [cpp] class Solution { public: int maxCount(int m, int n, vector<vector<int>>& ops) { int minRow = m, minCol = n; for (const auto &op : ops) { minRow = min(minRow, op[0]); minCol = min(minCol, op[1]); } return minRow*minCol; } }; [/cpp] 本代码提交AC，用时6MS。 ]]>

LeetCode Array Nesting A zero-indexed array A consisting of N different integers is given. The array contains all integers in the range [0, N – 1]. Sets S[K] for 0 <= K < N are defined as follows: S[K] = { A[K], A[A[K]], A[A[A[K]]], … }. Sets S[K] are finite for each K and should NOT contain duplicates. Write a function that given an array A consisting of N integers, return the size of the largest set S[K] for this array. Example 1:

Input: A = [5,4,0,3,1,6,2]
Output: 4
Explanation:
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.
One of the longest S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}

Note:

1. N is an integer within the range [1, 20,000].
2. The elements of A are all distinct.
3. Each element of array A is an integer within the range [0, N-1].

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求嵌套集合的最大长度。首先给定原数组A，嵌套集合为S[K] = { A[K], A[A[K]], A[A[A[K]]], … }，就是不断的把值当下标递归的取值，因为数组A中的值的范围也是0~n-1的，所以下标不会超出范围。 暴力方法就是求出每一个S[K]的大小，然后求最大值。但是仔细分析一个样例就会发现，很多S[K]集合是完全一样的，比如第一个样例中，S[0]和S[2]都等于{5，6，2，0}，因为他们构成了一个循环。所以我们可以创建一个visited数组，访问过就置1，只对哪些visited为0的下标开始递归。这样对于第一个样例，其实只需要3次递归就可以了，也就是下标到值的构成的图中环的个数：{5,6,2,0}，{3}，{1,4}。所以总的复杂度其实只有O(n)。 代码如下： [cpp] class Solution { private: int nest(vector<int> &nums, vector<int> &visited, int k) { int ans = 0; while (visited[k] == 0) { visited[k] = 1; ++ans; k = nums[k]; } return ans; } public: int arrayNesting(vector<int>& nums) { int n = nums.size(); vector<int> visited(n, 0); int ans = 0; for (int i = 0; i < nums.size(); ++i) { if (visited[i] == 0) { int cur = nest(nums, visited, i); ans = max(ans, cur); } } return ans; } }; [/cpp] 本代码提交AC，用时39MS。]]>

LeetCode Tag Validator Given a string representing a code snippet, you need to implement a tag validator to parse the code and return whether it is valid. A code snippet is valid if all the following rules hold:

1. The code must be wrapped in a valid closed tag. Otherwise, the code is invalid.
2. A closed tag (not necessarily valid) has exactly the following format : <TAG_NAME>TAG_CONTENT</TAG_NAME>. Among them, <TAG_NAME> is the start tag, and </TAG_NAME> is the end tag. The TAG_NAME in start and end tags should be the same. A closed tag is valid if and only if the TAG_NAME and TAG_CONTENT are valid.
3. A valid TAG_NAME only contain upper-case letters, and has length in range [1,9]. Otherwise, the TAG_NAME is invalid.
4. A valid TAG_CONTENT may contain other valid closed tags, cdata and any characters (see note1) EXCEPT unmatched <, unmatched start and end tag, and unmatched or closed tags with invalid TAG_NAME. Otherwise, the TAG_CONTENT is invalid.
5. A start tag is unmatched if no end tag exists with the same TAG_NAME, and vice versa. However, you also need to consider the issue of unbalanced when tags are nested.
6. A < is unmatched if you cannot find a subsequent >. And when you find a < or </, all the subsequent characters until the next > should be parsed as TAG_NAME (not necessarily valid).
7. The cdata has the following format : <![CDATA[CDATA_CONTENT]]>. The range of CDATA_CONTENT is defined as the characters between <![CDATA[ and the first subsequent]]>.
8. CDATA_CONTENT may contain any characters. The function of cdata is to forbid the validator to parse CDATA_CONTENT, so even it has some characters that can be parsed as tag (no matter valid or invalid), you should treat it as regular characters.

Valid Code Examples:

Input: "<DIV>This is the first line <![CDATA[<div>]]></DIV>"
Output: True
Explanation:
The code is wrapped in a closed tag : <DIV> and </DIV>.
The TAG_NAME is valid, the TAG_CONTENT consists of some characters and cdata.
Although CDATA_CONTENT has unmatched start tag with invalid TAG_NAME, it should be considered as plain text, not parsed as tag.
So TAG_CONTENT is valid, and then the code is valid. Thus return true.
Input: "<DIV>>>  ![cdata[]] <![CDATA[<div>]>]]>]]>>]</DIV>"
Output: True
Explanation:
We first separate the code into : start_tag|tag_content|end_tag.
start_tag -> "<DIV>"
end_tag -> "</DIV>"
tag_content could also be separated into : text1|cdata|text2.
text1 -> ">>  ![cdata[]] "
cdata -> "<![CDATA[<div>]>]]>", where the CDATA_CONTENT is "<div>]>"
text2 -> "]]>>]"
The reason why start_tag is NOT "<DIV>>>" is because of the rule 6.
The reason why cdata is NOT "<![CDATA[<div>]>]]>]]>" is because of the rule 7.

Invalid Code Examples:

Input: "<A>  <B> </A>   </B>"
Output: False
Explanation: Unbalanced. If "<A>" is closed, then "<B>" must be unmatched, and vice versa.
Input: "<DIV>  div tag is not closed  <DIV>"
Output: False
Input: "<DIV>  unmatched <  </DIV>"
Output: False
Input: "<DIV> closed tags with invalid tag name  <b>123</b> </DIV>"
Output: False
Input: "<DIV> unmatched tags with invalid tag name  </1234567890> and <CDATA[[]]>  </DIV>"
Output: False
Input: "<DIV>  unmatched start tag <B>  and unmatched end tag </C>  </DIV>"
Output: False

Note:

1. For simplicity, you could assume the input code (including the any characters mentioned above) only contain letters, digits, '<','>','/','!','[',']' and ' '.

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本题需要写一个检查简单html源代码标签是否匹配的程序。规则有很多，8条，比如tag长度必须在[1,9]，且必须是大写字母，还有<![CDATA[CDATA_CONTENT]]>这个东西，经常会在html源代码中看见。 这题在比赛的时候没想清楚，思路比较乱。比赛结束之后，看了讨论区的解答，感觉思路好清晰，借鉴过来。 需要检查的主要有两个部分，一是Tag标签匹配，二是cdata。对于cdata，只要找到，则可以过滤掉<![CDATA[和]]>之间的所有内容。对于Tag标签匹配，类似括号匹配，可以用栈来实现。 但是因为Tag和cdata的开头都有<，我们匹配的时候， 应该从特殊到一般，即首先考虑是<![CDATA，其次考虑是</，然后考虑是<，最后就是一般性的字符了。 当遇到<![CDATA时，找到对应的]]>，下标快速跳到]]>的下一位。 当遇到</时，说明遇到了一个结束Tag，此时判断Tag是否合法，如果合法，从栈中弹出开始Tag，判断开始Tag和结束Tag是否匹配。 当遇到<时，说明遇到了一个开始Tag，判断Tag是否合法，如果合法，则压栈。 如果是一般性的字符，下标直接向前进。 需要注意的是，如果i!=0但是栈为空，说明当前的字符不在某个Tag标签内，比如这个样例：<![CDATA[wahaha]]]><![CDATA[]> wahaha]]>，一上来就是cdata，这是不合法的，因为所有内容都应该在标签对内。 完整代码如下： [cpp] class Solution { private: bool checkTag(const string &tag) { if (tag.size() < 1 || tag.size() > 9)return false; for (const auto &c : tag) { if (!isupper(c))return false; } return true; } public: bool isValid(string code) { stack<string> stk; int i = 0; while (i < code.size()) { if (i > 0 && stk.empty())return false; if (code.substr(i, 9) == "<![CDATA[") { int end = code.find("]]>", i); if (end == string::npos)return false; i = end + 3; } else if (code.substr(i, 2) == "</") { int end = code.find(‘>’, i); if (end == string::npos)return false; string tag = code.substr(i + 2, end – i – 2); if (!checkTag(tag))return false; if (stk.empty() || stk.top() != tag)return false; else stk.pop(); i = end + 1; } else if (code[i] == ‘<‘) { int end = code.find(‘>’, i); if (end == string::npos)return false; string tag = code.substr(i + 1, end – i – 1); if (!checkTag(tag))return false; stk.push(tag); i = end + 1; } else { ++i; } } return stk.empty(); } }; [/cpp] 本代码提交AC，用时12MS。]]>

LeetCode Construct String from Binary Tree You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way. The null node needs to be represented by empty parenthesis pair “()”. And you need to omit all the empty parenthesis pairs that don’t affect the one-to-one mapping relationship between the string and the original binary tree. Example 1:

Input: Binary tree: [1,2,3,4]
       1
     /   \
    2     3
   /
  4
Output: "1(2(4))(3)"
Explanation: Originallay it needs to be "1(2(4)())(3()())",
but you need to omit all the unnecessary empty parenthesis pairs.
And it will be "1(2(4))(3)".

Example 2:

Input: Binary tree: [1,2,3,null,4]
       1
     /   \
    2     3
     \
      4
Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example,
except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

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本题要求把一棵二叉树转换为一个字符串，使用先序遍历的方法，并且左右子树需要用括号括起来，但是要保证从该字符串能唯一恢复回原来的二叉树，而且要去掉不必要的空括号。 首先明确先序遍历：根左右，不难。然后要删掉不必要的空括号，观察第二个样例，发现如果左孩子为空，则左孩子的空括号不能省，但是如果右孩子或者左右孩子都为空，则他们的空括号可以省略。所以分几种情况递归求解即可。 代码如下： [cpp] class Solution { public: string tree2str(TreeNode* t) { if (!t)return ""; if (!t->left && !t->right)return to_string(t->val); if (!t->left)return to_string(t->val) + "()(" + tree2str(t->right) + ")"; if (!t->right)return to_string(t->val) + "(" + tree2str(t->left) + ")"; return to_string(t->val) + "(" + tree2str(t->left) + ")(" + tree2str(t->right) + ")"; } }; [/cpp] 本代码提交AC，用时19MS。]]>

LeetCode Can Place Flowers Suppose you have a long flowerbed in which some of the plots are planted and some are not. However, flowers cannot be planted in adjacent plots – they would compete for water and both would die. Given a flowerbed (represented as an array containing 0 and 1, where 0 means empty and 1 means not empty), and a number n, return if n new flowers can be planted in it without violating the no-adjacent-flowers rule. Example 1:

Input: flowerbed = [1,0,0,0,1], n = 1
Output: True

Example 2:

Input: flowerbed = [1,0,0,0,1], n = 2
Output: False

Note:

1. The input array won’t violate no-adjacent-flowers rule.
2. The input array size is in the range of [1, 20000].
3. n is a non-negative integer which won’t exceed the input array size.

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今天第一次做Leetcode的weekly contest，AC了三题，最后一题看起来好繁琐，没A。 第一题，简单贪心即可，从第0个位置开始，检查每个位置的前后是否为1，如果不为1，说明该位置可以种花，填上1，最后统计新填上的1的个数和n的关系。 代码如下： [cpp] class Solution { public: bool canPlaceFlowers(vector<int>& flowerbed, int n) { int m = flowerbed.size(); for (int i = 0; i < m; ++i) { if (flowerbed[i] == 0) { if (i – 1 >= 0 && flowerbed[i – 1] == 1)continue; if (i + 1 < m&&flowerbed[i + 1] == 1)continue; flowerbed[i] = 1; –n; } } return n <= 0; } }; [/cpp] 本代码提交AC，用时26MS。]]>

241. Different Ways to Add Parentheses241. Different Ways to Add Parentheses

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.

Example 1:

Input: "2-1-1"
Output: [0, 2]
Explanation: 
((2-1)-1) = 0 
(2-(1-1)) = 2

Example 2:

Input: "2*3-4*5" Output: [-34, -14, -10, -10, 10] Explanation:  (2*(3-(4*5))) = -34  ((2*3)-(4*5)) = -14  ((2*(3-4))*5) = -10  (2*((3-4)*5)) = -10  (((2*3)-4)*5) = 10241. Different Ways to Add Parentheses

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给定一个包含+-*的运算字符串，求所有加括号的方案计算到的值。
采用分治法的思路，尝试在每个操作符分成两个部分，这就相当于分别把操作符左右两边括起来单独计算。然后left和right分开进行递归，最后merge。递归的终止条件是当字符串中不包含操作符时，直接返回这个字符串对应的数值。
分治法很经典的例子。
代码如下：

class Solution {
private:
    int operate(int a, int b, char c)
    {
        switch (c) {
        case ‘+’:
            return a + b;
        case ‘-‘:
            return a – b;
        case ‘*’:
            return a * b;
        }
    }
public:
    vector<int> diffWaysToCompute(string input)
    {
        int n = input.size();
        vector<int> ans;
        bool found = false;
        for (int i = 0; i < n; ++i) {
            if (input[i] == ‘+’ || input[i] == ‘-‘ || input[i] == ‘*’) {
                found = true;
                vector<int> lefts = diffWaysToCompute(input.substr(0, i));
                vector<int> rights = diffWaysToCompute(input.substr(i + 1));
                for (int j = 0; j < lefts.size(); ++j) {
                    for (int k = 0; k < rights.size(); ++k) {
                        ans.push_back(operate(lefts[j], rights[k], input[i]));
                    }
                }
            }
        }
        if (!found)
            return { atoi(input.c_str()) };
        return ans;
    }
};

本代码提交AC，用时3MS。