keep going!

amazon.cpp

@@ -3,47 +3,47 @@
 // O(n^2)时间，double for loop
 class Solution {
 public:
-    vector<int> twoSum(vector<int>& nums, int target) {
-        for(int outer=0; outer<nums.size()-1; outer++){
-            vector<int> result;
-            result.push_back(outer);
-            int temp = target;
-            temp -= nums[outer];
-            for(int inner=outer+1; inner<nums.size(); inner++){
-                if(temp-nums[inner] == 0){
-                    result.push_back(inner);
-                    return result;
-                }
-            }
+  vector<int> twoSum(vector<int>& nums, int target) {
+    for(int outer=0; outer<nums.size()-1; outer++){
+      vector<int> result;
+      result.push_back(outer);
+      int temp = target;
+      temp -= nums[outer];
+      for(int inner=outer+1; inner<nums.size(); inner++){
+        if(temp-nums[inner] == 0){
+          result.push_back(inner);
+          return result;
         }
-        vector<int> result;
-        return result;
+      }
     }
+    vector<int> result;
+    return result;
+  }
 };
 
 // solution 2
 // O(n)时间，using hash table to find
 class Solution {
 public:
-    vector<int> twoSum(vector<int>& nums, int target) {
-        // first is the rest, second is the indice
-        unordered_map<int, int> hash;
-        for(int i=0; i<nums.size(); i++){
-          int rest = target - nums[i];
-
-          auto iter = hash.find(rest);
-          if (iter != hash.end()) {
-            std::vector<int> result;
-            result.push_back(i);
-            result.push_back(iter->second);
-            return result;
-          }
-            hash[nums[i]] = i;
-        }
-
+  vector<int> twoSum(vector<int>& nums, int target) {
+    // first is the rest, second is the indice
+    unordered_map<int, int> hash;
+    for(int i=0; i<nums.size(); i++){
+      int rest = target - nums[i];
+
+      auto iter = hash.find(rest);
+      if (iter != hash.end()) {
         std::vector<int> result;
+        result.push_back(i);
+        result.push_back(iter->second);
         return result;
+      }
+      hash[nums[i]] = i;
     }
+
+    std::vector<int> result;
+    return result;
+  }
 };
 
 
@@ -53,45 +53,109 @@ public:
 // 然后依次判断左右高低，高的保留，低的跳过，左右筛选后再计算water值，直到最后计算出最大water
 class Solution {
 public:
-    int maxArea(vector<int>& height) {
-        int left = 0;
-        int right = height.size() -1;
-        int water = 0;
-
-        while(left < right){
-            water = max(water, ((right - left) * min(height[left], height[right])));
-            if(height[left] < height[right]){
-                left++;
-            }
-            else{
-                right--;
-            }
-        }
-
-        return water;
+  int maxArea(vector<int>& height) {
+    int left = 0;
+    int right = height.size() -1;
+    int water = 0;
+
+    while(left < right){
+      water = max(water, ((right - left) * min(height[left], height[right])));
+      if(height[left] < height[right]){
+        left++;
+      }
+      else{
+        right--;
+      }
     }
-};
-
-
-
-
-
-
 
+    return water;
+  }
+};
 
 
 
 
+// 3Sum Closest
+// almost same as 3 sum,但是在最后的三个if else语句中的判断有逻辑上的区别
+class Solution {
+public:
+  int threeSumClosest(vector<int>& nums, int target) {
+    sort(nums.begin(), nums.end());
+    int closest = INFINITY;
+    int result;
+    for (int outer = 0; outer < nums.size()-1; outer++) {
+      int left = outer+1;
+      int right = nums.size()-1;
+      while (left < right) {
+        int sum = nums[outer] + nums[left] + nums[right];
+        int diff = target - sum;
+        if (abs(diff) < closest) {
+            closest = abs(diff);
+            result = sum;
+        }
+        if (diff < 0) {
+         right--;
+        }
+        else if (diff > 0) {
+            left++;
+        }
+        else{
+          return result;
+        }
+      }
+    }
+    return result;
+  }
+};
 
 
 
+// Implement strStr()
+// KMP 算法的实现
+// step 1. pre-process两个string，建立最长prefix的array
+// step 2.若遇到不符，skip多少个由查ips表所知
+/*
+lps[i] = the longest proper prefix of pat[0..i]
+           which is also a suffix of pat[0..i].
 
+For the pattern “ABCDE”,
+pat = "AAAA"
+lps[] is [0, 0, 0, 0, 0]
 
+For the pattern “AABAACAABAA”,
+pat = "AAAA"
+lps[] is [0, 1, 0, 1, 2, 0, 1, 2, 3, 4, 5]
+*/
 
 
 
 
+// Rotate Image
+// 矩阵旋转：
+// 首先上下swap
+// 然后对称线swap
+// temp作为对角线终点变量（每一行到哪里停）
+/*
+ * clockwise rotate
+ * first reverse up to down, then swap the symmetry
+ * 1 2 3     7 8 9     7 4 1
+ * 4 5 6  => 4 5 6  => 8 5 2
+ * 7 8 9     1 2 3     9 6 3
+*/
+class Solution {
+public:
+    void rotate(vector<vector<int>>& matrix) {
+        swap(matrix.begin(), matrix.end());
 
+        int temp = 1;
+        for (int outer = 0; outer < matrix.size(); outer++) {
+          for (int inner = 0; inner < temp; inner++) {
+            swap(matrix[outer][inner], matrix[inner][outer]);
+          }
+          temp++;
+        }
+    }
+};