小修改

Author: soulmachine

C++/LeetCodet题解(C++版).pdf

@@ -409,8 +409,8 @@ \subsection{代码模板}
             const state_t state = current.front();
             current.pop();
             vector<state_t> new_states = state_extend(state, visited);
-            for (auto iter = new_states.begin();
-                    iter != new_states.end() && ! found; ++iter) {
+            for (auto iter = new_states.cbegin();
+                    iter != new_states.cend() && ! found; ++iter) {
                 const state_t new_state(*iter);
 
                 if (state_is_target(new_state)) {

C++/chapBFS.tex

@@ -1123,17 +1123,17 @@ \subsubsection{动规}
 public:
     bool wordBreak(string s, unordered_set<string> &dict) {
         // 长度为n的字符串有n+1个隔板
-        vector<bool> f(s.length() + 1, false);
-        f[0] = true;
-        for (int i = 1; i <= s.length(); ++i) {
+        vector<bool> f(s.size() + 1, false);
+        f[0] = true; // 空字符串
+        for (int i = 1; i <= s.size(); ++i) {
             for (int j = i - 1; j >= 0; --j) {
                 if (f[j] && dict.find(s.substr(j, i - j)) != dict.end()) {
                     f[i] = true;
                     break;
                 }
             }
         }
-        return f[s.length()];
+        return f[s.size()];
     }
 };
 \end{Code}
@@ -1176,9 +1176,9 @@ \subsubsection{代码}
         vector<bool> f(s.length() + 1, false);
         // path[i][j]为true，表示s[j, i)是一个合法单词，可以从j处切开
         // 第一行未用
-        vector<vector<bool> > prev(s.length()+1, vector<bool>(s.length()));
+        vector<vector<bool> > prev(s.length() + 1, vector<bool>(s.length()));
         f[0] = true;
-        for (int i = 1; i <= s.length(); ++i) {
+        for (size_t i = 1; i <= s.length(); ++i) {
             for (int j = i - 1; j >= 0; --j) {
                 if (f[j] && dict.find(s.substr(j, i - j)) != dict.end()) {
                     f[i] = true;
@@ -1198,12 +1198,12 @@ \subsubsection{代码}
             int cur, vector<string> &path, vector<string> &result) {
         if (cur == 0) {
             string tmp;
-            for (auto iter = path.rbegin(); iter != path.rend(); ++iter)
+            for (auto iter = path.crbegin(); iter != path.crend(); ++iter)
                 tmp += *iter + " ";
-            tmp.erase(tmp.end()-1);
+            tmp.erase(tmp.end() - 1);
             result.push_back(tmp);
         }
-        for (int i = 0; i < s.length(); ++i) {
+        for (size_t i = 0; i < s.size(); ++i) {
             if (prev[cur][i]) {
                 path.push_back(s.substr(i, cur - i));
                 gen_path(s, prev, i, path, result);

C++/chapDynamicProgramming.tex

@@ -1614,14 +1614,14 @@ \subsubsection{代码}
         vector<int> increment(n);
 
         // 左右各扫描一遍
-        for (int i = 0, inc = 1; i < n; i++)
-            if (i >= 1 && ratings[i] > ratings[i - 1])
+        for (int i = 1, inc = 1; i < n; i++)
+            if (ratings[i] > ratings[i - 1])
                 increment[i] = max(inc++, increment[i]);
             else
                 inc = 1;
 
-        for (int i = n - 1, inc = 1; i >= 0; i--)
-            if (i < n - 1 && ratings[i] > ratings[i + 1])
+        for (int i = n - 2, inc = 1; i >= 0; i--)
+            if (ratings[i] > ratings[i + 1])
                 increment[i] = max(inc++, increment[i]);
             else
                 inc = 1;

C++/chapLinearList.tex

@@ -596,7 +596,7 @@ \subsubsection{递归版}
 public:
     bool isMatch(const char *s, const char *p) {
         if (*p == '*') {
-            while (*p == '*') ++p;
+            while (*p == '*') ++p;  //skip continuous '*'
             if (*p == '\0') return true;
             while (*s != '\0' && !isMatch(s, p)) ++s;
 
@@ -613,7 +613,7 @@ \subsubsection{递归版}
 \subsubsection{迭代版}
 \begin{Code}
 // LeetCode, Wildcard Matching
-// 迭代版，时间复杂度O(n)，空间复杂度O(1)
+// 迭代版，时间复杂度O(n*m)，空间复杂度O(1)
 class Solution {
 public:
     bool isMatch(const char *s, const char *p) {
@@ -626,7 +626,7 @@ \subsubsection{迭代版}
             case '*':
                 star = true;
                 s = str, p = ptr;
-                while (*p == '*') p++;
+                while (*p == '*') p++;  //skip continuous '*'
                 if (*p == '\0') return true;
                 str = s - 1;
                 ptr = p - 1;
@@ -998,7 +998,7 @@ \subsubsection{代码}
         }
 
         vector<string> result;
-        for (auto it = group.begin(); it != group.end(); it++) {
+        for (auto it = group.cbegin(); it != group.cend(); it++) {
             if (it->second.size() > 1)
                 result.insert(result.end(), it->second.begin(), it->second.end());
         }