From 78a6fa6b008f330cdefdbbca9a58131c86a8dd32 Mon Sep 17 00:00:00 2001
From: Mrzhudky <1183260441@qq.com>
Date: Wed, 17 Oct 2018 22:29:36 +0800
Subject: [PATCH 1/4] =?UTF-8?q?018.=E5=9B=9B=E6=95=B0=E4=B9=8B=E5=92=8C[Ja?=
 =?UTF-8?q?va]?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 solution/018.4Sum/README.md     | 96 +++++++++++++++++++++++++++++++++
 solution/018.4Sum/Solution.java | 67 +++++++++++++++++++++++
 2 files changed, 163 insertions(+)
 create mode 100644 solution/018.4Sum/README.md
 create mode 100644 solution/018.4Sum/Solution.java

diff --git a/solution/018.4Sum/README.md b/solution/018.4Sum/README.md
new file mode 100644
index 0000000000000..43f996ecb1d24
--- /dev/null
+++ b/solution/018.4Sum/README.md
@@ -0,0 +1,96 @@
+## 四数之和
+### 题目描述
+
+给定一个包含 n 个整数的数组 nums 和一个目标值 target，
+判断 nums 中是否存在四个元素 a，b，c 和 d ，
+使得 a + b + c + d 的值与 target 相等？
+找出所有满足条件且不重复的四元组。
+
+注意：
+
+答案中不可以包含重复的四元组。
+
+示例：
+
+给定数组 nums = [1, 0, -1, 0, -2, 2]，和 target = 0。
+
+满足要求的四元组集合为：
+[
+  [-1,  0, 0, 1],
+  [-2, -1, 1, 2],
+  [-2,  0, 0, 2]
+]
+
+
+### 解法
+先将4Sum问题，转换为3Sum问题，再转换为2Sum问题.
+
+```java
+class Solution {
+    public List<List<Integer>> fourSum(int[] nums, int target) {
+                
+		List<List<Integer>> re = new ArrayList<>();
+        if(nums == null || nums.length<4){
+            return  re;
+        }
+        Arrays.sort(nums);
+        for (int i = 0; i < nums.length-3; i++) {
+		
+            //当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
+            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target){
+                break;
+            }
+            // 当 nums[i] 对应的最大组合都小于 target 时， nums[i] 的其他组合必然也小于 target
+            if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target){
+                continue;
+            }
+
+            int firstNum = nums[i];
+            for (int j = i+1; j < nums.length-2; j++) {
+
+                // nums[j] 过大时，与nums[i]过大同理
+                if(nums[i]+nums[j]+nums[j+1]+nums[j+2] > target){
+                    break;
+                }
+                // nums[j] 过小时，与nums[i]过小同理
+                if(nums[i]+nums[j]+nums[nums.length-2] + nums[nums.length-1] < target){
+                    continue;
+                }
+
+                int twoSum = target - nums[i]-nums[j];
+                int l = j+1;
+                int k = nums.length-1;
+                while (l<k){
+                    int tempSum = nums[l] + nums[k];
+                    if(tempSum == twoSum){
+                        ArrayList<Integer> oneGroup = new ArrayList<>(4);
+                        oneGroup.add(nums[i]);
+                        oneGroup.add(nums[j]);
+                        oneGroup.add(nums[l++]);
+                        oneGroup.add(nums[k--]);
+                        re.add(oneGroup);
+                        while (l<nums.length && l<k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)){
+                            l++;k--;
+                        }
+                    }
+                    else if(tempSum<twoSum){
+                        l++;
+                    }
+                    else{
+                        k--;
+                    }
+                }
+                // 跳过重复项
+                while ((j<nums.length-2)&&(twoSum + nums[i] + nums[j+1] == target)){
+                    j++;
+                }
+            }
+            // 跳过重复项
+            while (i<nums.length-3 &&(nums[i+1] == firstNum) ){
+                i++;
+            }
+        }
+        return re;
+    }
+}
+```
\ No newline at end of file
diff --git a/solution/018.4Sum/Solution.java b/solution/018.4Sum/Solution.java
new file mode 100644
index 0000000000000..c0e18db999c5b
--- /dev/null
+++ b/solution/018.4Sum/Solution.java
@@ -0,0 +1,67 @@
+class Solution {
+    public List<List<Integer>> fourSum(int[] nums, int target) {
+                
+		List<List<Integer>> re = new ArrayList<>();
+        if(nums == null || nums.length<4){
+            return  re;
+        }
+        Arrays.sort(nums);
+        for (int i = 0; i < nums.length-3; i++) {
+		
+            //当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
+            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target){
+                break;
+            }
+            // 当 nums[i] 对应的最大组合都小于 target 时， nums[i] 的其他组合必然也小于 target
+            if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target){
+                continue;
+            }
+
+            int firstNum = nums[i];
+            for (int j = i+1; j < nums.length-2; j++) {
+
+                // nums[j] 过大时，与nums[i]过大同理
+                if(nums[i]+nums[j]+nums[j+1]+nums[j+2] > target){
+                    break;
+                }
+                // nums[j] 过小时，与nums[i]过小同理
+                if(nums[i]+nums[j]+nums[nums.length-2] + nums[nums.length-1] < target){
+                    continue;
+                }
+
+                int twoSum = target - nums[i]-nums[j];
+                int l = j+1;
+                int k = nums.length-1;
+                while (l<k){
+                    int tempSum = nums[l] + nums[k];
+                    if(tempSum == twoSum){
+                        ArrayList<Integer> oneGroup = new ArrayList<>(4);
+                        oneGroup.add(nums[i]);
+                        oneGroup.add(nums[j]);
+                        oneGroup.add(nums[l++]);
+                        oneGroup.add(nums[k--]);
+                        re.add(oneGroup);
+                        while (l<nums.length && l<k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)){
+                            l++;k--;
+                        }
+                    }
+                    else if(tempSum<twoSum){
+                        l++;
+                    }
+                    else{
+                        k--;
+                    }
+                }
+                // 跳过重复项
+                while ((j<nums.length-2)&&(twoSum + nums[i] + nums[j+1] == target)){
+                    j++;
+                }
+            }
+            // 跳过重复项
+            while (i<nums.length-3 &&(nums[i+1] == firstNum) ){
+                i++;
+            }
+        }
+        return re;
+    }
+}
\ No newline at end of file

From dbac5bebdc4fe5ee19a3d77b7e2a32b1a8e9596d Mon Sep 17 00:00:00 2001
From: Mrzhudky <1183260441@qq.com>
Date: Thu, 18 Oct 2018 10:07:13 +0800
Subject: [PATCH 2/4] Add solution 018[Java]

---
 solution/018.4Sum/README.md     | 35 ++++++++++++++++++---------------
 solution/018.4Sum/Solution.java | 30 ++++++++++++++--------------
 2 files changed, 34 insertions(+), 31 deletions(-)

diff --git a/solution/018.4Sum/README.md b/solution/018.4Sum/README.md
index 43f996ecb1d24..ce67d8bf1326b 100644
--- a/solution/018.4Sum/README.md
+++ b/solution/018.4Sum/README.md
@@ -23,70 +23,73 @@
 
 
 ### 解法
-先将4Sum问题，转换为3Sum问题，再转换为2Sum问题.
+1、将数组排序
+2、先假设确定一个数 nums[i] 将 4Sum 问题转换为 3Sum 问题；
+3、再假设确定一个数将 3Sum 问题转换为 2Sum 问题；
+4、对排序数组，用首尾指针向中间靠拢的思路寻找满足 target 的 nums[l] 和 nums[k]
 
 ```java
 class Solution {
     public List<List<Integer>> fourSum(int[] nums, int target) {
                 
 		List<List<Integer>> re = new ArrayList<>();
-        if(nums == null || nums.length<4){
+        if(nums == null || nums.length<4) {
             return  re;
         }
         Arrays.sort(nums);
         for (int i = 0; i < nums.length-3; i++) {
 		
-            //当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
-            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target){
+            // 当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
+            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target) {
                 break;
             }
             // 当 nums[i] 对应的最大组合都小于 target 时， nums[i] 的其他组合必然也小于 target
-            if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target){
+            if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target) {
                 continue;
             }
 
             int firstNum = nums[i];
             for (int j = i+1; j < nums.length-2; j++) {
 
-                // nums[j] 过大时，与nums[i]过大同理
-                if(nums[i]+nums[j]+nums[j+1]+nums[j+2] > target){
+                // nums[j] 过大时，与 nums[i] 过大同理
+                if(nums[i] + nums[j] + nums[j+1] + nums[j+2] > target) {
                     break;
                 }
-                // nums[j] 过小时，与nums[i]过小同理
-                if(nums[i]+nums[j]+nums[nums.length-2] + nums[nums.length-1] < target){
+                // nums[j] 过小时，与 nums[i] 过小同理
+                if(nums[i] + nums[j] + nums[nums.length-2] + nums[nums.length-1] < target) {
                     continue;
                 }
 
                 int twoSum = target - nums[i]-nums[j];
                 int l = j+1;
                 int k = nums.length-1;
-                while (l<k){
+                while (l<k) {
                     int tempSum = nums[l] + nums[k];
-                    if(tempSum == twoSum){
+                    if(tempSum == twoSum) {
                         ArrayList<Integer> oneGroup = new ArrayList<>(4);
                         oneGroup.add(nums[i]);
                         oneGroup.add(nums[j]);
                         oneGroup.add(nums[l++]);
                         oneGroup.add(nums[k--]);
                         re.add(oneGroup);
-                        while (l<nums.length && l<k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)){
+                        while (l < nums.length && l < k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)) {
                             l++;k--;
                         }
                     }
-                    else if(tempSum<twoSum){
+                    else if(tempSum < twoSum) {
                         l++;
                     }
-                    else{
+                    else {
                         k--;
                     }
                 }
                 // 跳过重复项
-                while ((j<nums.length-2)&&(twoSum + nums[i] + nums[j+1] == target)){
+                while ((j < nums.length - 2) && (twoSum + nums[i] + nums[j+1] == target)) {
                     j++;
                 }
             }
             // 跳过重复项
-            while (i<nums.length-3 &&(nums[i+1] == firstNum) ){
+            while (i < nums.length-3 && nums[i+1] == firstNum){
                 i++;
             }
         }
diff --git a/solution/018.4Sum/Solution.java b/solution/018.4Sum/Solution.java
index c0e18db999c5b..0df98a1a610d2 100644
--- a/solution/018.4Sum/Solution.java
+++ b/solution/018.4Sum/Solution.java
@@ -2,63 +2,63 @@ class Solution {
     public List<List<Integer>> fourSum(int[] nums, int target) {
                 
 		List<List<Integer>> re = new ArrayList<>();
-        if(nums == null || nums.length<4){
+        if(nums == null || nums.length<4) {
             return  re;
         }
         Arrays.sort(nums);
         for (int i = 0; i < nums.length-3; i++) {
 		
-            //当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
-            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target){
+            // 当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
+            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target) {
                 break;
             }
             // 当 nums[i] 对应的最大组合都小于 target 时， nums[i] 的其他组合必然也小于 target
-            if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target){
+            if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target) {
                 continue;
             }
 
             int firstNum = nums[i];
             for (int j = i+1; j < nums.length-2; j++) {
 
-                // nums[j] 过大时，与nums[i]过大同理
-                if(nums[i]+nums[j]+nums[j+1]+nums[j+2] > target){
+                // nums[j] 过大时，与 nums[i] 过大同理
+                if(nums[i] + nums[j] + nums[j+1] + nums[j+2] > target) {
                     break;
                 }
-                // nums[j] 过小时，与nums[i]过小同理
-                if(nums[i]+nums[j]+nums[nums.length-2] + nums[nums.length-1] < target){
+                // nums[j] 过小时，与 nums[i] 过小同理
+                if(nums[i] + nums[j] + nums[nums.length-2] + nums[nums.length-1] < target) {
                     continue;
                 }
 
                 int twoSum = target - nums[i]-nums[j];
                 int l = j+1;
                 int k = nums.length-1;
-                while (l<k){
+                while (l<k) {
                     int tempSum = nums[l] + nums[k];
-                    if(tempSum == twoSum){
+                    if(tempSum == twoSum) {
                         ArrayList<Integer> oneGroup = new ArrayList<>(4);
                         oneGroup.add(nums[i]);
                         oneGroup.add(nums[j]);
                         oneGroup.add(nums[l++]);
                         oneGroup.add(nums[k--]);
                         re.add(oneGroup);
-                        while (l<nums.length && l<k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)){
+                        while (l < nums.length && l < k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)) {
                             l++;k--;
                         }
                     }
-                    else if(tempSum<twoSum){
+                    else if(tempSum < twoSum) {
                         l++;
                     }
-                    else{
+                    else {
                         k--;
                     }
                 }
                 // 跳过重复项
-                while ((j<nums.length-2)&&(twoSum + nums[i] + nums[j+1] == target)){
+                while ((j < nums.length - 2) && (twoSum + nums[i] + nums[j+1] == target)) {
                     j++;
                 }
             }
             // 跳过重复项
-            while (i<nums.length-3 &&(nums[i+1] == firstNum) ){
+            while (i < nums.length-3 && nums[i+1] == firstNum){
                 i++;
             }
         }

From 102b19eb6de5de586d6b588633818efa535d6d09 Mon Sep 17 00:00:00 2001
From: Mrzhudky <1183260441@qq.com>
Date: Thu, 18 Oct 2018 10:14:23 +0800
Subject: [PATCH 3/4] Update format and solution description For Solution
 018[Java]

---
 solution/018.4Sum/README.md     | 20 ++++++++++----------
 solution/018.4Sum/Solution.java | 20 ++++++++++----------
 2 files changed, 20 insertions(+), 20 deletions(-)

diff --git a/solution/018.4Sum/README.md b/solution/018.4Sum/README.md
index ce67d8bf1326b..7fca0a5c2a1a6 100644
--- a/solution/018.4Sum/README.md
+++ b/solution/018.4Sum/README.md
@@ -33,23 +33,23 @@ class Solution {
     public List<List<Integer>> fourSum(int[] nums, int target) {
                 
 		List<List<Integer>> re = new ArrayList<>();
-        if(nums == null || nums.length<4) {
+        if(nums == null || nums.length < 4) {
             return  re;
         }
         Arrays.sort(nums);
-        for (int i = 0; i < nums.length-3; i++) {
+        for (int i = 0; i < nums.length - 3; i++) {
 		
             // 当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
-            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target) {
+            if(nums[i] + nums[i+1] + nums[i+2] + nums[i+3] > target) {
                 break;
             }
             // 当 nums[i] 对应的最大组合都小于 target 时， nums[i] 的其他组合必然也小于 target
-            if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target) {
+            if(nums[i] + nums[nums.length-3] + nums[nums.length-2] + nums[nums.length-1] < target) {
                 continue;
             }
 
             int firstNum = nums[i];
-            for (int j = i+1; j < nums.length-2; j++) {
+            for (int j = i + 1; j < nums.length - 2; j++) {
 
                 // nums[j] 过大时，与 nums[i] 过大同理
                 if(nums[i] + nums[j] + nums[j+1] + nums[j+2] > target) {
@@ -60,10 +60,10 @@ class Solution {
                     continue;
                 }
 
-                int twoSum = target - nums[i]-nums[j];
-                int l = j+1;
-                int k = nums.length-1;
-                while (l<k) {
+                int twoSum = target - nums[i] - nums[j];
+                int l = j + 1;
+                int k = nums.length - 1;
+                while (l < k) {
                     int tempSum = nums[l] + nums[k];
                     if(tempSum == twoSum) {
                         ArrayList<Integer> oneGroup = new ArrayList<>(4);
@@ -89,7 +89,7 @@ class Solution {
                 }
             }
             // 跳过重复项
-            while (i < nums.length-3 && nums[i+1] == firstNum){
+            while (i < nums.length - 3 && nums[i+1] == firstNum){
                 i++;
             }
         }
diff --git a/solution/018.4Sum/Solution.java b/solution/018.4Sum/Solution.java
index 0df98a1a610d2..2d93cc423c4df 100644
--- a/solution/018.4Sum/Solution.java
+++ b/solution/018.4Sum/Solution.java
@@ -2,23 +2,23 @@ class Solution {
     public List<List<Integer>> fourSum(int[] nums, int target) {
                 
 		List<List<Integer>> re = new ArrayList<>();
-        if(nums == null || nums.length<4) {
+        if(nums == null || nums.length < 4) {
             return  re;
         }
         Arrays.sort(nums);
-        for (int i = 0; i < nums.length-3; i++) {
+        for (int i = 0; i < nums.length - 3; i++) {
 		
             // 当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
-            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target) {
+            if(nums[i] + nums[i+1] + nums[i+2] + nums[i+3] > target) {
                 break;
             }
             // 当 nums[i] 对应的最大组合都小于 target 时， nums[i] 的其他组合必然也小于 target
-            if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target) {
+            if(nums[i] + nums[nums.length-3] + nums[nums.length-2] + nums[nums.length-1] < target) {
                 continue;
             }
 
             int firstNum = nums[i];
-            for (int j = i+1; j < nums.length-2; j++) {
+            for (int j = i + 1; j < nums.length - 2; j++) {
 
                 // nums[j] 过大时，与 nums[i] 过大同理
                 if(nums[i] + nums[j] + nums[j+1] + nums[j+2] > target) {
@@ -29,10 +29,10 @@ public List<List<Integer>> fourSum(int[] nums, int target) {
                     continue;
                 }
 
-                int twoSum = target - nums[i]-nums[j];
-                int l = j+1;
-                int k = nums.length-1;
-                while (l<k) {
+                int twoSum = target - nums[i] - nums[j];
+                int l = j + 1;
+                int k = nums.length - 1;
+                while (l < k) {
                     int tempSum = nums[l] + nums[k];
                     if(tempSum == twoSum) {
                         ArrayList<Integer> oneGroup = new ArrayList<>(4);
@@ -58,7 +58,7 @@ else if(tempSum < twoSum) {
                 }
             }
             // 跳过重复项
-            while (i < nums.length-3 && nums[i+1] == firstNum){
+            while (i < nums.length - 3 && nums[i+1] == firstNum){
                 i++;
             }
         }

From 408e8e9ca22e6027e4b69c181fae3a27c3dcd25d Mon Sep 17 00:00:00 2001
From: Mrzhudky <1183260441@qq.com>
Date: Thu, 18 Oct 2018 10:19:38 +0800
Subject: [PATCH 4/4] Repair list display problem

---
 solution/018.4Sum/README.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/solution/018.4Sum/README.md b/solution/018.4Sum/README.md
index 7fca0a5c2a1a6..3894e0b6e6393 100644
--- a/solution/018.4Sum/README.md
+++ b/solution/018.4Sum/README.md
@@ -23,10 +23,10 @@
 
 
 ### 解法
-1、将数组排序
-2、先假设确定一个数 nums[i] 将 4Sum 问题转换为 3Sum 问题；
-3、再假设确定一个数将 3Sum 问题转换为 2Sum 问题；
-4、对排序数组，用首尾指针向中间靠拢的思路寻找满足 target 的 nums[l] 和 nums[k]
+1. 将数组排序
+2. 先假设确定一个数 nums[i] 将 4Sum 问题转换为 3Sum 问题；
+3. 再假设确定一个数将 3Sum 问题转换为 2Sum 问题；
+4. 对排序数组，用首尾指针向中间靠拢的思路寻找满足 target 的 nums[l] 和 nums[k]
 
 ```java
 class Solution {