From 22819ff61b6ee93dce393e38124c2921ef4e865d Mon Sep 17 00:00:00 2001
From: liu13 <1099976891@qq.com>
Date: Wed, 10 Apr 2019 21:23:22 +0800
Subject: [PATCH] 20190410

---
 code/lc148.java | 43 +++++++++++++++++++++++++++++++++++++++++++
 code/lc51.java  |  2 +-
 code/lc52.java  | 41 +++++++++++++++++++++++++++++++++++++++++
 code/lc923.java | 19 +------------------
 4 files changed, 86 insertions(+), 19 deletions(-)
 create mode 100644 code/lc52.java

diff --git a/code/lc148.java b/code/lc148.java
index 24c54a1..7ab9390 100644
--- a/code/lc148.java
+++ b/code/lc148.java
@@ -1,10 +1,17 @@
 package code;
+
+import java.util.ArrayList;
+import java.util.List;
+
 /*
  * 148. Sort List
  * 题意：链表排序
  * 难度：Medium
  * 分类：Linked List, Sort
  * 思路：快慢指针把链表分成两半，在merge两个链表
+ *      快排方法自己不会写，记下思路
+ *      快排尝试直接以ListNode互相交换位置，节点间的指向会乱掉的，当递归子情况的时候会修改指针，父方法不知道子调用做了哪些操作
+ *      https://www.cnblogs.com/morethink/p/8452914.html
  * Tips：空间复杂度不是O(1)的，但是几个高票答案都是这样写的，面试给出这样的代码应该也够了
  */
 public class lc148 {
@@ -52,4 +59,40 @@ public class lc148 {
         }
         return head.next;
     }
+
+    public ListNode sortList2(ListNode head) {  //链表快排
+        //采用快速排序
+        quickSort(head, null);
+        return head;
+    }
+    public void quickSort(ListNode head, ListNode end) {
+        if (head != end) {
+            ListNode node = partion(head, end);
+            quickSort(head, node);
+            quickSort(node.next, end);
+        }
+    }
+
+    public ListNode partion(ListNode head, ListNode end) {
+        ListNode p1 = head, p2 = head.next;
+
+        //走到末尾才停
+        while (p2 != end) { //p1与p2间都是大于pivot的数
+            if (p2.val < head.val) {    //lc922 类似的思想， 把小于的值放到该放的位置上
+                p1 = p1.next;
+
+                int temp = p1.val;
+                p1.val = p2.val;
+                p2.val = temp;
+            }
+            p2 = p2.next;
+        }
+
+        //与pivot交换下位置
+        int temp = p1.val;
+        p1.val = head.val;
+        head.val = temp;
+
+        return p1;  //返回
+    }
 }
diff --git a/code/lc51.java b/code/lc51.java
index 9c2dcba..97acd76 100644
--- a/code/lc51.java
+++ b/code/lc51.java
@@ -5,7 +5,7 @@ package code;
  * 难度：Hard
  * 分类：Backtracking
  * 思路：回溯+判断，注意怎么判断两个斜线方向
- * Tips：
+ * Tips：lc52
  */
 import java.util.ArrayList;
 import java.util.HashSet;
diff --git a/code/lc52.java b/code/lc52.java
new file mode 100644
index 0000000..114083b
--- /dev/null
+++ b/code/lc52.java
@@ -0,0 +1,41 @@
+package code;
+
+import java.util.HashSet;
+/*
+ * 52. N-Queens II
+ * 题意：8皇后问题
+ * 难度：Hard
+ * 分类：Backtracking
+ * 思路：和 lc51 一样，输出变为种类数，反而简单了
+ * Tips：
+ */
+public class lc52 {
+    HashSet<String> hs = new HashSet();
+    int res = 0;
+    public int totalNQueens(int n) {
+        dfs(n, 0);
+        return res;
+    }
+
+    public void dfs(int n, int row){
+        if(row==n){
+            res++;
+        }
+        for (int i = 0; i < n ; i++) {
+            if(isValid(row, i)){
+                int a = row+i;  // row+col 作为key
+                int b = row-i;  // row-col 作为key
+                hs.add("row"+row); hs.add("col"+i); hs.add("k1"+a); hs.add("k2"+b);
+                dfs(n, row+1);
+                hs.remove("row"+row); hs.remove("col"+i); hs.remove("k1"+a); hs.remove("k2"+b); //别忘了删掉
+            }
+        }
+    }
+
+    public boolean isValid(int row, int col){
+        int a = row+col;
+        int b = row-col;
+        if(hs.contains("row"+row)||hs.contains("col"+col)||hs.contains("k1"+a)||hs.contains("k2"+b)) return false;
+        return true;
+    }
+}
diff --git a/code/lc923.java b/code/lc923.java
index 44c0a32..726cbc1 100644
--- a/code/lc923.java
+++ b/code/lc923.java
@@ -9,7 +9,7 @@ import java.util.HashMap;
  * 分类：Two Pointers
  * 思路：由于本题只要求给出多少种Int值，所以不一定非要用3Sum的思路，有很多更简答的方法
  *      3种思路
- * Tips：
+ * Tips：lc15, lc16, lc923
  */
 public class lc923 {
     public int threeSumMulti(int[] A, int target) {
@@ -53,21 +53,4 @@ public class lc923 {
         return res;
     }
 
-    public int threeSumMulti3(int[] A, int target) {    //直接数学计算
-        long[] c = new long[101];
-        for (int a : A) c[a]++;
-        long res = 0;
-        for (int i = 0; i <= 100; i++)  // 题目给了值不超过100
-            for (int j = i; j <= 100; j++) {
-                int k = target - i - j;
-                if (k > 100 || k < 0) continue;
-                if (i == j && j == k)
-                    res += c[i] * (c[i] - 1) * (c[i] - 2) / 6;
-                else if (i == j && j != k)
-                    res += c[i] * (c[i] - 1) / 2 * c[k];
-                else if (j < k)
-                    res += c[i] * c[j] * c[k];
-            }
-        return (int)(res % (1e9 + 7));
-    }
 }
-- 
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