Add solution 015

Author: yanglbme

README.md

@@ -38,6 +38,7 @@ Complete solutions to Leetcode problems, updated daily.
 | # | Title | Tags |
 |---|---|---|
 | 002 | [Add Two Numbers](https://github.com/yanglbme/leetcode/tree/master/solution/002.Add%20Two%20Numbers) | `Linked List`, `Math` |
+| 015 | [3Sum](https://github.com/yanglbme/leetcode/tree/master/solution/015.3Sum) | `Array`, `Two Pointers` |
 | 019 | [Remove Nth Node From End of List](https://github.com/yanglbme/leetcode/tree/master/solution/019.Remove%20Nth%20Node%20From%20End%20of%20List) | `Linked List`, `Two Pointers` |
 | 024 | [Swap Nodes in Pairs](https://github.com/yanglbme/leetcode/tree/master/solution/024.Swap%20Nodes%20in%20Pairs) | `Linked List` |
 | 046 | [Permutations](https://github.com/yanglbme/leetcode/tree/master/solution/046.Permutations) | `Backtracking` |

solution/015.3Sum/README.md

@@ -0,0 +1,67 @@
+## 三数之和
+### 题目描述
+
+给定一个包含 n 个整数的数组 nums，判断 nums 中是否存在三个元素 a，b，c ，使得 a + b + c = 0 ？找出所有满足条件且不重复的三元组。
+
+注意：答案中不可以包含重复的三元组。
+```
+例如, 给定数组 nums = [-1, 0, 1, 2, -1, -4]，
+
+满足要求的三元组集合为：
+[
+  [-1, 0, 1],
+  [-1, -1, 2]
+]
+```
+
+### 解法
+先对数组进行排序，遍历数组，固定第一个数i。利用两个指针 p, q 分别指示 i+1, n-1。如果三数之和为0，移动 p, q；如果大于 0，左移 q；如果小于 0，右移 p。遍历到 nums[i] > 0 时，退出循环。
+
+```java
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        List<List<Integer>> list = new ArrayList<>();
+        Arrays.sort(nums);
+        int n = nums.length;
+        if (n < 3) {
+            return list;
+        }
+        int p = 0;
+        int q = 0;
+        for (int i = 0; i < n - 2; ++i) {
+            if (nums[i] > 0) {
+                break;
+            }
+            if (i > 0 && nums[i] == nums[i - 1]) {
+                continue;
+            }
+            
+            p = i + 1;
+            q = n - 1;
+            
+            while (p < q) {
+                int val = nums[p] + nums[q] + nums[i];
+                if (val == 0) {
+                    list.add(Arrays.asList(nums[i], nums[p], nums[q]));
+                    ++p;
+                    while (p < q && nums[p] == nums[p - 1]) {
+                        ++p;
+                    }
+                    --q;
+                    while (p < q && nums[q] == nums[q + 1]) {
+                        --q;
+                    }
+                    
+                } else {
+                    q = val > 0 ? q - 1 : q;
+                    p = val < 0 ? p + 1 : p;
+                }
+                
+            }  
+        }
+        return list;
+        
+    }
+    
+}
+```

solution/015.3Sum/Solution.java

@@ -0,0 +1,46 @@
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        List<List<Integer>> list = new ArrayList<>();
+        Arrays.sort(nums);
+        int n = nums.length;
+        if (n < 3) {
+            return list;
+        }
+        int p = 0;
+        int q = 0;
+        for (int i = 0; i < n - 2; ++i) {
+            if (nums[i] > 0) {
+                break;
+            }
+            if (i > 0 && nums[i] == nums[i - 1]) {
+                continue;
+            }
+            
+            p = i + 1;
+            q = n - 1;
+            
+            while (p < q) {
+                int val = nums[p] + nums[q] + nums[i];
+                if (val == 0) {
+                    list.add(Arrays.asList(nums[i], nums[p], nums[q]));
+                    ++p;
+                    while (p < q && nums[p] == nums[p - 1]) {
+                        ++p;
+                    }
+                    --q;
+                    while (p < q && nums[q] == nums[q + 1]) {
+                        --q;
+                    }
+                    
+                } else {
+                    q = val > 0 ? q - 1 : q;
+                    p = val < 0 ? p + 1 : p;
+                }
+                
+            }  
+        }
+        return list;
+        
+    }
+    
+}