Problem

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Idea:

Time complexity O(n^2)

Space complexity O(n^2)

Solutions:

C++

// Author: Huahua

class Solution {

public:

bool wordBreak(string s, vector<string>& wordDict) {

// Create a hashset of words for fast query.

unordered_set<string> dict(wordDict.cbegin(), wordDict.cend());

// Query the answer of the original string.

return wordBreak(s, dict);

}

bool wordBreak(const string& s, const unordered_set<string>& dict) {

// In memory, directly return.

if(mem_.count(s)) return mem_[s];

// Whole string is a word, memorize and return.

if(dict.count(s)) return mem_[s]=true;

// Try every break point.

for(int j=1;j<s.length();j++) {

const string left = s.substr(0,j);

const string right = s.substr(j);

// Find the solution for s.

if(dict.count(right) && wordBreak(left, dict))

return mem_[s]=true;

}

// No solution for s, memorize and return.

return mem_[s]=false;

}

private:

unordered_map<string, bool> mem_;

};

C++ V2 without using dict. Updated: 1/9/2018

// Author: Huahua

// Running time: 9 ms

class Solution {

public:

bool wordBreak(string s, vector<string>& wordDict) {

// Mark evert word as breakable.

for (const string& word : wordDict)

mem_.emplace(word, true);

// Query the answer of the original string.

return wordBreak(s);

}

bool wordBreak(const string& s) {

// In memory, directly return.

if (mem_.count(s)) return mem_[s];

// Try every break point.

for (int j = 1; j < s.length(); j++) {

auto it = mem_.find(s.substr(j));

// Find the solution for s.

if (it != mem_.end() && it->second && wordBreak(s.substr(0, j)))

return mem_[s] = true;

}

// No solution for s, memorize and return.

return mem_[s] = false;

}

private:

unordered_map<string, bool> mem_;

};

Java

// Author: Huahua

// Runtime: 13 ms

class Solution {

public boolean wordBreak(String s, List<String> wordDict) {

Set<String> dict = new HashSet<String>(wordDict);

Map<String, Boolean> mem = new HashMap<String, Boolean>();

return wordBreak(s, mem, dict);

}

private boolean wordBreak(String s,

Map<String, Boolean> mem,

Set<String> dict) {

if (mem.containsKey(s)) return mem.get(s);

if (dict.contains(s)) {

mem.put(s, true);

return true;

}

for (int i = 1; i < s.length(); ++i) {

if (dict.contains(s.substring(i)) && wordBreak(s.substring(0, i), mem, dict)) {

mem.put(s, true);

return true;

}

mem.put(s, false);

return false;

}

Python

"""

Author: Huahua

Runtime: 42 ms

"""

class Solution(object):

def wordBreak(self, s, wordDict):

def canBreak(s, m, wordDict):

if s in m: return m[s]

if s in wordDict:

m[s] = True

return True

for i in range(1, len(s)):

r = s[i:]

if r in wordDict and canBreak(s[0:i], m, wordDict):

m[s] = True

return True

m[s] = False

return False

return canBreak(s, {}, set(wordDict))

Problem:

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example 1:

Input:     1         1
          / \       / \
         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true

Example 2:

Input:     1         1
          /           \
         2             2

        [1,2],     [1,null,2]

Output: false

Example 3:

Input:     1         1
          / \       / \
         2   1     1   2

        [1,2,1],   [1,1,2]

Output: false

Solution: Recursion

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

bool isSameTree(TreeNode* p, TreeNode* q) {

// Both are emtry: same

if (!p && !q) return true;

// One is emtry: not the Same

if (!p || !q) return false;

// Both are not emptry, compare the root value

if (p->val != q->val) return false;

// Compare left-subtree and right-subtree recursively.

return isSameTree(p->left, q->left)

&& isSameTree(p->right, q->right);

}

};

Java

// Author: Huahua

class Solution {

public boolean isSameTree(TreeNode p, TreeNode q) {

if (p == null && q == null) return true;

if (p == null || q == null) return false;

return p.val == q.val && isSameTree(p.left, q.left) && isSameTree(p.right, q.right);

}

Python3

"""

Author: Huahua

"""

class Solution:

def isSameTree(self, p, q):

if not p and not q: return True

if not p or not q: return False

return all((p.val == q.val,

self.isSameTree(p.left, q.left),

self.isSameTree(p.right, q.right)))

Follow Up

花花酱 LeetCode 572. Subtree of Another Tree

Posts published in September 2017

花花酱 Leetcode 139. Word Break

Problem

Idea:

Solutions:

花花酱 Leetcode 100. Same Tree

Problem:

Example 1:

Example 2:

Example 3:

Solution: Recursion

C++

Java

Python3

Follow Up