May 19, 2018

Author: Xuemeng Zhang

139_Word_Break.cpp

@@ -0,0 +1,48 @@
+/*
+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.
+
+Note:
+
+The same word in the dictionary may be reused multiple times in the segmentation.
+You may assume the dictionary does not contain duplicate words.
+Example 1:
+
+Input: s = "leetcode", wordDict = ["leet", "code"]
+Output: true
+Explanation: Return true because "leetcode" can be segmented as "leet code".
+Example 2:
+
+Input: s = "applepenapple", wordDict = ["apple", "pen"]
+Output: true
+Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
+             Note that you are allowed to reuse a dictionary word.
+Example 3:
+
+Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
+Output: false
+*/
+#include <vector>
+#include <string>
+using namespace std;
+
+class Solution {
+public:
+    bool wordBreak(string s, vector<string>& wordDict) {
+		int *p = new int[s.length()+1];
+		for(int i=0; i<=(int)s.length(); ++i) p[i] = 0;
+		p[0] = 1;
+		for(int i=1; i<=(int)s.length(); ++i){
+			for(vector<string>::iterator iter = wordDict.begin(); \
+					iter != wordDict.end(); ++iter){
+				int j = i - iter->size();
+				if( (0<=j) && p[j] && (s.substr(j, iter->size())==*iter) ){
+					p[i] = 1;
+					break;
+				}
+			}
+		}
+		bool result = p[s.length()];
+		delete[] p;
+		return result;
+    }
+};

242_Valid_Anagram.cpp

@@ -0,0 +1,41 @@
+/*
+Given two strings s and t , write a function to determine if t is an anagram of s.
+
+Example 1:
+
+Input: s = "anagram", t = "nagaram"
+Output: true
+Example 2:
+
+Input: s = "rat", t = "car"
+Output: false
+Note:
+You may assume the string contains only lowercase alphabets.
+
+Follow up:
+What if the inputs contain unicode characters? How would you adapt your solution to such case?
+
+*/
+#include <string>
+#include <vector>
+using namespace std;
+
+class Solution {
+public:
+    bool isAnagram(string s, string t) {
+		//check the length
+		if(s.length()!=t.length()) return false;
+		//build counting table
+		int counts[26] = {0};
+		for(size_t i=0; i<s.length(); ++i){
+			counts[s[i]-'a']++;
+			counts[t[i]-'a']--;
+		}
+		//check if counting table only contains zeros
+		for(size_t i=0; i<26; ++i){
+			if(counts[i]) return false;
+		}
+		//return
+		return true;
+    }
+};