string

Author: vegetabledogg

3-Longest-Substring-Without-Repeating-Characters.cpp

@@ -0,0 +1,58 @@
+/*
+ * 第一种是O(n^3)的解法，没有什么难点
+ */
+class Solution {
+public:
+    int lengthOfLongestSubstring(string s) {
+        int max = 0, i, j, k, flag;
+        for(i = 0; i < s.size(); i++){
+            for(j = 1; j < s.size() - i; j++){
+                flag = 0;
+                for(k = i; k < i + j; k++){
+                    if(s[k] == s[i + j]){
+                        flag = 1;
+                        break;
+                    }
+                }
+                if(flag == 1){
+                    break;
+                }
+            }
+            if(j > max){
+                max = j;
+            }
+        }
+        return max;
+    }
+};
+
+/*
+ * 2n steps，C++的set用红黑树实现，每一步还需考虑库函数的时间复杂度
+ */
+class Solution {
+public:
+    int lengthOfLongestSubstring(string s) {
+        if(s.size() == 0){
+            return 0;
+        }
+        int max = 1, i = 0, j = 1;
+        set<char> myset;
+        myset.insert(s[0]);
+        while(j < s.size() - i){
+            if(myset.find(s[i + j]) == myset.end()){
+                myset.insert(s[i + j]);
+                j++;
+                if(j > max){
+                    max = j;
+                }
+            }
+            else{
+                myset.erase(s[i]);
+                i++;
+                j--;
+            }
+        }
+        return max;
+    }
+};
+

5-Longest-Palindromic-Substring.cpp

@@ -0,0 +1,85 @@
+class Solution {
+public:
+    string longestPalindrome(string s) {
+        int max = 0, i, j, length = s.size();
+        string maxString = "";
+        if(length == 0){
+            return maxString;
+        }
+        for(i = 0; i < length; i++){
+            for(j = 1; i - j >= 0 && i + j < length; j++){
+                if(s[i - j] != s[i + j]){
+                    break;
+                }
+            }
+            if(max < j * 2 - 1){
+                max = j * 2 - 1;
+                maxString = s.substr(i - j + 1, max);
+            }
+        }
+        for(i = 0; i < length - 1; i++){
+            if(s[i] == s[i + 1]){
+                for(j = 1; i - j >= 0 && i + 1 + j < length; j++){
+                    if(s[i - j] != s[i + 1 + j]){
+                        break;
+                    }
+                }
+                if(max < j * 2){
+                    max = j * 2;
+                    maxString = s.substr(i - j + 1, max);
+                }
+            }
+        }
+        return maxString;
+    }
+};
+
+/*
+ * Manacher's Algorithm
+ */
+class Solution {
+public:
+    string longestPalindrome(string s) {
+        string str = "#";
+        int length = s.size(), i, j;
+        for(i = 0; i < length; i++){
+            str = str + s[i] + "#";
+        }
+        length = str.size();
+        int *len = new int[length];
+        int p = 0, max = 0, po = 0;
+        for(i = 0; i < length; i++){
+            if(i <= p){
+                if(len[2 * po - i] <= p - i + 1){
+                    len[i] = len[2 * po - i];
+                }
+                else{
+                    len[i] = p - i + 1;
+                }
+            }
+            else{
+                len[i] = 1;
+            }
+            for(j = len[i]; i - j >= 0 && i + j < length; j++){
+                if(str[i - j] != str[i + j]){
+                    break;
+                }
+            }
+            len[i] = j;
+            if(p < i + len[i] - 1){
+                po = i;
+                p = i + len[i] - 1;
+            }
+            if(len[max] < j){
+                max = i;
+            }
+        }
+        string res = "";
+        int left = max - len[max] + 1, right = max + len[max] - 1;
+        for(i = left + 1; i <= right - 1; i += 2){
+            res += str[i];
+        }
+        return res;
+    }
+};
+