fix 3Sum TLE

Author: soulmachine

C++/chapLinearList.tex

@@ -547,78 +547,37 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 // LeetCode, 3Sum
-// 先排序，然后左右夹逼，时间复杂度O(n^2)，空间复杂度O(1)
+// 先排序，然后左右夹逼，注意跳过重复的数，时间复杂度O(n^2)，空间复杂度O(1)
 class Solution {
-public:
+    public:
     vector<vector<int>> threeSum(vector<int>& num) {
         vector<vector<int>> result;
         if (num.size() < 3) return result;
         sort(num.begin(), num.end());
         const int target = 0;
-
+		
         auto last = num.end();
-        for (auto a = num.begin(); a < prev(last, 2); ++a) {
-            auto b = next(a);
-            auto c = prev(last);
-            while (b < c) {
-                if (*a + *b + *c < target) {
-                    ++b;
-                } else if (*a + *b + *c > target) {
-                    --c;
+        for (auto i = num.begin(); i < last-2; ++i) {
+            auto j = i+1;
+            if (i > num.begin() && *i == *(i-1)) continue;
+            auto k = last-1;
+            while (j < k) {
+                if (*i + *j + *k < target) {
+                    ++j;
+                    while(*j == *(j - 1) && j < k) ++j;
+                } else if (*i + *j + *k > target) {
+                    --k;
+                    while(*k == *(k + 1) && j < k) --k;
                 } else {
-                    result.push_back({ *a, *b, *c });
-                    ++b;
-                    --c;
-                }
-            }
-        }
-        sort(result.begin(), result.end());
-        result.erase(unique(result.begin(), result.end()), result.end());
-        return result;
-    }
-};
-\end{Code}
-
-\subsubsection{分析}
-先排序，然后左右夹逼，复杂度 $O(n^2)$。
-需要返回没有duplicate的结果。每次寻找新结果就排除那些可能重复的数字，外层i如果与之前一样则进行下一个，
-同理对里层夹逼的两个数也是这样来避免。
-\subsubsection{代码}
-\begin{Code}
-class Solution {
-public:
-vector<vector<int> > threeSum(vector<int> &num){
-    sort(num.begin(), num.end());
-    vector<vector<int> > result;
-    if(num.size() < 3) return result;
-    for(int i = 0; i < num.size() - 2; ++i){
-        if(i  && num[i] == num[i - 1]) continue;
-        int j = i + 1;
-        int k = num.size() - 1;
-        while(j < k){
-            if(num[i] + num[j] + num[k] == 0){
-                result.push_back({num[i], num[j], num[k]});
-                ++j, --k;
-                while(num[j] == num[j - 1] && num[k] == num[k + 1] && j < k){
-                    ++j, --k;
-                }
-            }
-            else if(num[i] + num[j] + num[k] < 0){
-                ++j;
-                while(num[j] == num[j - 1] && j < k){
+                    result.push_back({ *i, *j, *k });
                     ++j;
-                }
-            }
-            else{
-                --k;
-                while(num[k] == num[k + 1] && j < k){
                     --k;
+                    while(*j == *(j - 1) && *k == *(k + 1) && j < k) ++j;
                 }
             }
         }
+        return result;
     }
-    return result;
-  }
 };
 \end{Code}