199/208

Author: MaskRay

README.md

@@ -2,7 +2,7 @@
 
 [LeetCode solutions](http://maskray.me/blog/2014-06-29-leetcode-solutions) gives some thoughts on selected problems.
 
-Solved 196/205 problems.
+Solved 199/208 problems.
 
 ## Database
 
@@ -12,6 +12,9 @@ See [database.md](database.md)
 
 | # | Title | Solution |
 |---| ----- | -------- |
+|224|[Basic Calculator](https://leetcode.com/problems/basic-calculator/)|[basic-calculator.cc](basic-calculator.cc)|
+|223|[Rectangle Area](https://leetcode.com/problems/rectangle-area/)|[rectangle-area.cc](rectangle-area.cc)|
+|222|[Count Complete Tree Nodes](https://leetcode.com/problems/count-complete-tree-nodes/)|[count-complete-tree-nodes.cc](count-complete-tree-nodes.cc)|
 |221|[Maximal Square](https://leetcode.com/problems/maximal-square/)|[maximal-square.cc](maximal-square.cc)|
 |220|[Contains Duplicate III](https://leetcode.com/problems/contains-duplicate-iii/)|[contains-duplicate-iii.cc](contains-duplicate-iii.cc)|
 |219|[Contains Duplicate II](https://leetcode.com/problems/contains-duplicate-ii/)|[contains-duplicate-ii.cc](contains-duplicate-ii.cc)|

basic-calculator.cc

@@ -0,0 +1,38 @@
+// Basic Calculator
+// operator-precedence parser
+class Solution
+{
+public:
+  int calculate(string s) {
+    stack<int> xs;
+    stack<char> ops;
+    map<char, int> isp, icp;
+    isp['\0'] = 0; isp['+'] = 3; isp['-'] = 3; isp['('] = 1;
+    icp['\0'] = 0; icp['+'] = 2; icp['-'] = 2; icp['('] = 4; icp[')'] = 1;
+    ops.push('\0');
+    for (size_t i = 0; i <= s.size(); ) {
+      char op = i == s.size() ? '\0' : s[i];
+      if (op == ' ')
+        i++;
+      else if (isdigit(op)) {
+        int x = 0;
+        do x = x*10+s[i]-'0';
+        while (isdigit(s[++i]));
+        xs.push(x); // shift
+      } else {
+        i++;
+        int x, y;
+        for (; isp[ops.top()] > icp[op]; ops.pop()) {
+          y = xs.top(); xs.pop();
+          x = xs.top(); xs.pop();
+          xs.push(ops.top() == '+' ? x+y : x-y); // reduce, X '+' Y => X+Y  or  X '-' Y => X-Y
+        }
+        if (isp[ops.top()] == icp[op]) // reduce, '(' X ')' => X  or  '\0' X '\0' => X
+          ops.pop();
+        else
+          ops.push(op); // shift
+      }
+    }
+    return xs.top();
+  }
+};

count-complete-tree-nodes.cc

@@ -0,0 +1,12 @@
+// Count Complete Tree Nodes
+class Solution {
+public:
+  int countNodes(TreeNode* root) {
+    if (! root) return 0;
+    int dl = 0, dr = 0;
+    TreeNode *l = root, *r = root;
+    while (l) l = l->left, dl++;
+    while (r) r = r->right, dr++;
+    return dl == dr ? (1 << dl) - 1 : countNodes(root->left) + countNodes(root->right) + 1;
+  }
+};

evaluate-reverse-polish-notation.cc

@@ -1,30 +1,30 @@
 // Evaluate Reverse Polish Notation
 class Solution {
 public:
-    int evalRPN(vector<string> &tokens) {
-      stack<int> st;
-      int x, y;
-      for (auto &s: tokens)
-        if ('0' <= s[0] && s[0] <= '9' || s[0] == '-' && s.size() > 1)
-          st.push(atoi(s.c_str()));
-        else {
-          y = st.top(); st.pop();
-          x = st.top(); st.pop();
-          switch (s[0]) {
-          case '+':
-            st.push(x+y);
-            break;
-          case '-':
-            st.push(x-y);
-            break;
-          case '*':
-            st.push(x*y);
-            break;
-          case '/':
-            st.push(x/y);
-            break;
-          }
+  int evalRPN(vector<string> &tokens) {
+    stack<int> st;
+    int x, y;
+    for (auto &s: tokens)
+      if ('0' <= s[0] && s[0] <= '9' || s[0] == '-' && s.size() > 1)
+        st.push(atoi(s.c_str()));
+      else {
+        y = st.top(); st.pop();
+        x = st.top(); st.pop();
+        switch (s[0]) {
+        case '+':
+          st.push(x+y);
+          break;
+        case '-':
+          st.push(x-y);
+          break;
+        case '*':
+          st.push(x*y);
+          break;
+        case '/':
+          st.push(x/y);
+          break;
         }
-      return st.top();
-    }
+      }
+    return st.top();
+  }
 };

palindrome-partitioning-ii.cc

@@ -1,28 +1,19 @@
 // Palindrome Partitioning II
 #define FOR(i, a, b) for (int i = (a); i < (b); i++)
-#define ROF(i, a, b) for (int i = (b); --i >= (a); )
+#define REP(i, n) FOR(i, 0, n)
 
 class Solution {
 public:
   int minCut(string s) {
-    if (s.empty()) return 0;
     int n = s.size();
-    vector<vector<bool>> f(n, vector<bool>(n));
-    vector<int> g(n+1);
-    g[n] = 0;
-    ROF(i, 0, n) {
-      f[i][i] = true;
-      if (i+1 < n)
-        f[i][i+1] = s[i] == s[i+1];
-      FOR(j, i+2, n)
-        if (f[i+1][j-1] && s[i] == s[j])
-          f[i][j] = true;
-      g[i] = n;
-      FOR(j, i, n)
-        if (f[i][j])
-          g[i] = min(g[i], g[j+1]+1);
+    vector<int> f(n+1);
+    iota(f.begin(), f.end(), -1);
+    REP(i, n) {
+      for (int j = 0; i-j >= 0 && i+j < n && s[i-j] == s[i+j]; j++)
+        f[i+j+1] = min(f[i+j+1], f[i-j]+1);
+      for (int j = 1; i-j+1 >= 0 && i+j < n && s[i-j+1] == s[i+j]; j++)
+        f[i+j+1] = min(f[i+j+1], f[i-j+1]+1);
     }
-    return g[0]-1;
+    return f[n];
   }
 };
-

rectangle-area.cc

@@ -0,0 +1,13 @@
+// Rectangle Area
+// 存在給出的兩個頂點不是左下和右上的測試資料：-2, -2, 2, 2, -3, 3, -4, 4
+class Solution {
+public:
+  int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
+    if (A > C) swap(A, C);
+    if (B > D) swap(B, D);
+    if (E > G) swap(E, G);
+    if (F > H) swap(F, H);
+    int x = max(A, E), y = max(B, F), xx = min(C, G), yy = min(D, H);
+    return (C-A)*(D-B) + (G-E)*(H-F) - max(xx-x, 0) * max(yy-y, 0);
+  }
+};

repeated-dna-sequences.cc

@@ -1,4 +1,24 @@
 // Repeated DNA Sequences
+
+// rolling hash
+
+class Solution {
+public:
+  vector<string> findRepeatedDnaSequences(string a) {
+    unsigned i = 0, n = a.size(), t = 0;
+    unordered_map<unsigned, unsigned> m;
+    vector<string> r;
+    while (i < min(9u, n))
+      t = t<<3 | a[i++]&7;
+    while (i < n)
+      if (m[t = t<<3 & 0x3fffffff | a[i++]&7]++ == 1)
+        r.push_back(a.substr(i-10, 10));
+    return r;
+  }
+};
+
+// suffix array
+
 #define REP(i, n) FOR(i, 0, n)
 #define FOR(i, a, b) for (int i = (a); i < (b); i++)
 #define ROF(i, a, b) for (int i = (b); --i >= (a); )

shell.md

@@ -0,0 +1,23 @@
+## Tenth Line
+
+```bash
+sed -n 10p file.txt
+```
+
+## Transpose File
+
+```bash
+awk '{for(i=1;i<=NF;i++) {if(s[i]=="") s[i]=$i; else s[i]=s[i]" "$i}} END {for(i=1;s[i]!="";i++) print s[i]; }' file.txt
+```
+
+## Valid Phone Numbers
+
+```bash
+grep -P '^(\d{3}-|\(\d{3}\) )\d{3}-\d{4}$' file.txt
+```
+
+## Word Frequency
+
+```bash
+tr -s ' ' '\n' < words.txt | sort | uniq -c | sort -nr | awk '{print $2" "$1}'
+```

sort-list.cc

@@ -1,4 +1,7 @@
 // Sort List
+
+// natural merge sort
+
 class Solution {
 public:
   ListNode *sortList(ListNode *head) {

two-sum.cc

@@ -2,13 +2,11 @@
 class Solution {
 public:
     vector<int> twoSum(vector<int> &a, int s) {
-      size_t i = 0, j = 1;
       vector<int> r(a.size());
       iota(r.begin(), r.end(), 0);
       sort(r.begin(), r.end(), [&](int x, int y) { return a[x] < a[y]; });
-      while (j+1 < a.size() && a[r[i]]+a[r[j+1]] <= s) j++;
-      for (; i < j; i++) {
-        while (j > 0 && a[r[i]]+a[r[j]] > s) j--;
+      for (size_t i = 0, j = a.size()-1; i < j; i++) {
+        while (j > i+1 && a[r[i]]+a[r[j]] > s) j--;
         if (a[r[i]]+a[r[j]] == s) {
           int x = r[i], y = r[j];
           r.clear();