refactor

C++/chapStackAndQueue.tex

@@ -104,28 +104,26 @@ \subsubsection{使用栈}
 \subsubsection{Dynamic Programming, One Pass}
 \begin{Code}
 // LeetCode, Longest Valid Parenthese
-// 时间复杂度O(n)，空间复杂度O(1)
-// @author 一只杰森(http://weibo.com/2197839961)
+// 时间复杂度O(n)，空间复杂度O(n)
+// @author 一只杰森(http://weibo.com/wjson)
 class Solution {
 public:
     int longestValidParentheses(string s) {
         if (s.empty()) return 0;
-        int ret = 0;
         int* d = new int[s.size()];
         d[s.size() - 1] = 0;
+        int ret = 0;
         for (int i = s.size() - 2; i >= 0; --i) {
-            if (s[i] == ')') d[i] = 0;
-            else {
-                int match = i + d[i + 1] + 1;  // case: "((...))"
-                if (match < s.size() && s[match] == ')') {  // found matching ')'
-                    d[i] = match - i + 1;
-                    if (match + 1 < s.size()) d[i] += d[match + 1];  // more valid sequence after j: "((...))(...)"
-                } else {
-                    d[i] = 0;  // no matching ')'
-                }
+            int match = i + d[i + 1] + 1;
+            if (s[i] == '(' && match < s.size() && s[match] == ')') {
+                d[i] = d[i + 1] + 2;
+                if (match + 1 < s.size()) d[i] += d[match + 1];
+            } else {
+                d[i] = 0;
             }
             ret = max(ret, d[i]);
         }
+        delete[] d;
         return ret;
     }
 };