旧代码初步整理完成

Author: Richard-coder

Leetcode/array/3Sum/readme.md

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+# 15. 3Sum
+
+tags： Array, Two Pointers
+
+---
+
+## 题目原文
+Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
+
+Note: The solution set must not contain duplicate triplets.
+
+For example, given array S = [-1, 0, 1, 2, -1, -4],
+
+A solution set is:
+
+    [
+      [-1, 0, 1],
+      [-1, -1, 2]
+    ]
+
+## 题目大意
+给定一个整型数组num， 找出这个数组中满足这个条件的所有数字： num[i]+num[j]+num[k] = 0. 并且所有
+的答案是要和其他不同的， 也就是说两个相同的答案是不被接受的。
+题目的两点要求：
+1. 每个答案组里面的三个数字是要从大到小排列起来的。
+2. 每个答案不可以和其他的答案相同。 
+
+
+
+## 解题思路
+### 网上分析1
+题目分析：
+
+1. 每一个答案数组triplet中的元素是要求升序排列的。
+2. 不能包含重复的答案数组。
+
+解题思路：
+1. 根据第一点要求： 因为要求每个答案数组中的元素是升序排列的， 所以在开头我们要对数组进行排
+序。
+2. 根据第二点要求： 因为不能包含重复的答案数组， 所以我们要在代码里面做一切去掉重复的操作， 对
+于数组， 这样的操作是相同的。 最开始我做leetcode的时候是把所有满足条件的答案数组存起来， 之后
+再用map进行处理， 感觉那样太麻烦了， 所以这次给出的答案是不需要额外空间的。
+
+时间复杂度分析：
+对于这道题， 因为是要找三个元素， 所以怎样都要O(n2)的时间复杂度， 目前我没有想出来O(n)时间复杂度
+的解法。
+
+归根结底， 其实这是two pointers的想法， 定位其中两个指针， 根据和的大小来移动另外一个。 解题中要
+注意的就是一些细节问题。 好了， 上代码吧。
+ 
+### 网上分析2
+the key idea is the same as the TwoSum problem. When we fix the 1st number,the 2nd and 3rd number can be found following the same reasoning as TwoSum.
+
+The only difference is that, the TwoSum problem of LEETCODE has a unique solution.However, in ThreeSum, we have multiple duplicate solutions that can be found. Most ofthe OLE errors happened here because you could've ended up with a solution with so manyduplicates.
+
+The naive solution for the duplicates will be using the STL methods like below :
+
+std::sort(res.begin(), res.end());
+
+res.erase(unique(res.begin(), res.end()), res.end());
+
+But according to my submissions, this way will cause you double your time consumingalmostly.
+A better approach is that, to jump over the number which has been scanned, no matter itis part of some solution or not.
+
+If the three numbers formed a solution, we can safely ignore all the duplicates of them.
+We can do this to all the three numbers such that we can remove the duplicates.
+Here's my AC C++ Code:
+
+### 我的分析：
+
+主要是先将原数组排序，并充分利用排序好的数组的特性。去除重复答案的方法是，先按index从小到大确定第一个元素，在按（bengin=index）++和（end）--的顺序确定第二个元素和第三个元素，当begin=begin+1是跳过，end同理，第一个元素的所有可能循环完成后，index++之前也要判断index是否等于index+1。总而言之就是利用排序号数组特性保证每个元素都各不相同。先确定答案中最小的数，防止答案中出现重复的组合
+
+
+
+## 代码
+### [c++代码](./src/cpp/3Sum.cpp)
+```c++
+class Solution {
+public:
+    vector<vector<int>> threeSum(vector<int>& nums) {
+        std::sort(nums.begin(),nums.end());
+        vector<vector<int >> res;
+        for(int i=0;i<nums.size();i++){
+            int target=-nums[i];
+            int front=i+1;
+            int back=nums.size()-1;
+            while(front<back){
+                int sum=nums[front]+nums[back];
+                if(sum>target)
+                    back--;
+                else if(sum<target)
+                    front++;
+                else{
+                    vector<int> tri(3,0);
+                    tri[0]=nums[i];
+                    tri[1]=nums[front];
+                    tri[2]=nums[back];
+                    res.push_back(tri);
+                    
+                    while(front<back&&nums[front]==tri[1]) front++;
+                    while(front<back&&nums[back]==tri[2])  back--;
+                    
+
+                }
+                while(i+1<nums.size()&&nums[i+1]==nums[i]) i++;
+            }
+        }
+        return res;
+    }
+};
+```
+
+### [python代码](./src/python/3Sum.py)
+```python
+cimport collections
+
+
+class Solution(object):
+        def threeSum(self, nums):
+                """
+:type nums: List[int]
+:rtype: List[List[int]]
+"""
+                nums, result, i = sorted(nums), [], 0
+                while i < len(nums) - 2:
+                        if i == 0 or nums[i] != nums[i - 1]:
+                                j, k = i + 1, len(nums) - 1
+                                while j < k:
+                                        if nums[i] + nums[j] + nums[k] < 0:
+                                                j += 1
+                                        elif nums[i] + nums[j] + nums[k] > 0:
+                                                k -= 1
+                                        else:
+                                                result.append([nums[i], nums[j], nums[k]])
+                                                j, k = j + 1, k - 1
+                                                while j < k and nums[j] == nums[j - 1]:
+                                                        j += 1
+                                                while j < k and nums[k] == nums[k + 1]:
+                                                        k -= 1
+                        i += 1
+                return result
+
+        def threeSum2(self, nums):
+                """
+:type nums: List[int]
+:rtype: List[List[int]]
+"""
+                d = collections.Counter(nums)
+                nums_2 = [x[0] for x in d.items() if x[1] > 1]
+                nums_new = sorted([x[0] for x in d.items()])
+                rtn = [[0, 0, 0]] if d[0] >= 3 else []
+                for i, j in enumerate(nums_new):
+                        if j <= 0:
+                                numss2 = nums_new[i + 1:]
+                                for x, y in enumerate(numss2):
+                                        if 0 - j - y in [j, y] and 0 - j - y in nums_2:
+                                                if sorted([j, y, 0 - j - y]) not in rtn:
+                                                        rtn.append(sorted([j, y, 0 - j - y]))
+                                        if 0 - j - y not in [j, y] and 0 - j - y in nums_new:
+                                                if sorted([j, y, 0 - j - y]) not in rtn:
+                                                        rtn.append(sorted([j, y, 0 - j - y]))
+                return rtn
+
+if __name__ == '__main__':
+        result = Solution().threeSum([-1, 0, 1, 2, -1, -4])
+        print result
+
+```

Leetcode/array/3Sum/src/cpp/3Sum.cpp

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+class Solution {
+public:
+    vector<vector<int>> threeSum(vector<int>& nums) {
+        std::sort(nums.begin(),nums.end());
+        vector<vector<int >> res;
+        for(int i=0;i<nums.size();i++){
+            int target=-nums[i];
+            int front=i+1;
+            int back=nums.size()-1;
+            while(front<back){
+                int sum=nums[front]+nums[back];
+                if(sum>target)
+                    back--;
+                else if(sum<target)
+                    front++;
+                else{
+                    vector<int> tri(3,0);
+                    tri[0]=nums[i];
+                    tri[1]=nums[front];
+                    tri[2]=nums[back];
+                    res.push_back(tri);
+                    
+                    while(front<back&&nums[front]==tri[1]) front++;
+                    while(front<back&&nums[back]==tri[2])  back--;
+                    
+
+                }
+                while(i+1<nums.size()&&nums[i+1]==nums[i]) i++;
+            }
+        }
+        return res;
+    }
+};

Leetcode/array/3Sum/src/python/3Sum.py

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+import collections
+
+
+class Solution(object):
+        def threeSum(self, nums):
+                """
+:type nums: List[int]
+:rtype: List[List[int]]
+"""
+                nums, result, i = sorted(nums), [], 0
+                while i < len(nums) - 2:
+                        if i == 0 or nums[i] != nums[i - 1]:
+                                j, k = i + 1, len(nums) - 1
+                                while j < k:
+                                        if nums[i] + nums[j] + nums[k] < 0:
+                                                j += 1
+                                        elif nums[i] + nums[j] + nums[k] > 0:
+                                                k -= 1
+                                        else:
+                                                result.append([nums[i], nums[j], nums[k]])
+                                                j, k = j + 1, k - 1
+                                                while j < k and nums[j] == nums[j - 1]:
+                                                        j += 1
+                                                while j < k and nums[k] == nums[k + 1]:
+                                                        k -= 1
+                        i += 1
+                return result
+
+        def threeSum2(self, nums):
+                """
+:type nums: List[int]
+:rtype: List[List[int]]
+"""
+                d = collections.Counter(nums)
+                nums_2 = [x[0] for x in d.items() if x[1] > 1]
+                nums_new = sorted([x[0] for x in d.items()])
+                rtn = [[0, 0, 0]] if d[0] >= 3 else []
+                for i, j in enumerate(nums_new):
+                        if j <= 0:
+                                numss2 = nums_new[i + 1:]
+                                for x, y in enumerate(numss2):
+                                        if 0 - j - y in [j, y] and 0 - j - y in nums_2:
+                                                if sorted([j, y, 0 - j - y]) not in rtn:
+                                                        rtn.append(sorted([j, y, 0 - j - y]))
+                                        if 0 - j - y not in [j, y] and 0 - j - y in nums_new:
+                                                if sorted([j, y, 0 - j - y]) not in rtn:
+                                                        rtn.append(sorted([j, y, 0 - j - y]))
+                return rtn
+
+if __name__ == '__main__':
+        result = Solution().threeSum([-1, 0, 1, 2, -1, -4])
+        print result

Leetcode/array/MergeSortedArray/readme.md

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+88. Merge Sorted Array
+
+tags： Array
+
+---
+
+## 题目原文
+Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.
+
+Note:
+
+You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.
+## 题目大意
+nums1和nums2是两个排序好的数组，把两个数组合并放入nums1并排好序。
+
+
+## 解题思路
+A和B都已经是排好序的数组， 我们只需要从后往前比较就可以了。
+
+因为A有足够的空间容纳A + B， 我们使用游标i指向m + n - 1， 也就是最大数值存放的地方， 从后往前遍历A， B， 谁大就放到i这里， 同时递减i。
+
+
+
+## 代码
+### [c++代码](./src/cpp/MergeSortedArray.cpp)
+```c++
+class Solution {
+public:
+    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
+        int idx1=m-1;
+        int idx2=n-1;
+        //int idx=m+n-1;
+        for (int idx=m+n-1;idx>=0;idx--){
+            if(idx1>=0&&idx2>=0){
+                if (nums1[idx1]>=nums2[idx2])
+                    nums1[idx]=nums1[idx1--];
+                else
+                    nums1[idx]=nums2[idx2--];
+            }else{
+                if (idx1>=0)
+                    nums1[idx]=nums1[idx1--];
+                    
+                else
+                    nums1[idx]=nums2[idx2--];
+            }
+        }
+    }
+};
+```
+
+### [python代码](./src/python/MergeSortedArray.py)
+```python
+class Solution:
+        # @param A a list of integers
+        # @param m an integer, length of A
+        # @param B a list of integers
+        # @param n an integer, length of B
+        # @return nothing
+        def merge(self, A, m, B, n):
+                last, i, j = m + n - 1, m - 1, n - 1
+                
+                while i >= 0 and j >= 0:
+                        if A[i] > B[j]:
+                                A[last] = A[i]
+                                last, i = last - 1, i - 1
+                        else:
+                                A[last] = B[j]
+                                last, j = last - 1, j - 1
+                
+                while j >= 0:
+                                A[last] = B[j]
+                                last, j = last - 1, j - 1
+
+if __name__ == "__main__":
+        A = [1, 3, 5, 0, 0, 0, 0]
+        B = [2, 4, 6, 7]
+        Solution().merge(A, 3, B, 4)
+        print A
+
+
+```

Leetcode/array/MergeSortedArray/src/cpp/MergeSortedArray.cpp

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+class Solution {
+public:
+    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
+        int idx1=m-1;
+        int idx2=n-1;
+        //int idx=m+n-1;
+        for (int idx=m+n-1;idx>=0;idx--){
+            if(idx1>=0&&idx2>=0){
+                if (nums1[idx1]>=nums2[idx2])
+                    nums1[idx]=nums1[idx1--];
+                else
+                    nums1[idx]=nums2[idx2--];
+            }else{
+                if (idx1>=0)
+                    nums1[idx]=nums1[idx1--];
+                    
+                else
+                    nums1[idx]=nums2[idx2--];
+            }
+        }
+    }
+};

Leetcode/array/MergeSortedArray/src/python/MergeSortedArray.py

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+class Solution:
+        # @param A a list of integers
+        # @param m an integer, length of A
+        # @param B a list of integers
+        # @param n an integer, length of B
+        # @return nothing
+        def merge(self, A, m, B, n):
+                last, i, j = m + n - 1, m - 1, n - 1
+                
+                while i >= 0 and j >= 0:
+                        if A[i] > B[j]:
+                                A[last] = A[i]
+                                last, i = last - 1, i - 1
+                        else:
+                                A[last] = B[j]
+                                last, j = last - 1, j - 1
+                
+                while j >= 0:
+                                A[last] = B[j]
+                                last, j = last - 1, j - 1
+
+if __name__ == "__main__":
+        A = [1, 3, 5, 0, 0, 0, 0]
+        B = [2, 4, 6, 7]
+        Solution().merge(A, 3, B, 4)
+        print A
+