From abfdac6690b7ebcd0121d067c163126a9602327c Mon Sep 17 00:00:00 2001 From: liu13 <1099976891@qq.com> Date: Thu, 11 Jul 2019 11:08:05 +0800 Subject: [PATCH] 20190711 --- code/lc148.java | 4 ++-- code/lc153.java | 19 ++++++------------- code/lc206.java | 1 + code/lc236.java | 3 ++- code/lc25.java | 39 +++++++++++++++++++++++++++++++++++++++ code/lc300.java | 2 +- code/lc315.java | 2 +- code/lc72.java | 2 +- code/lc75.java | 2 +- code/lc912.java | 39 +++++++++++++++++++++++++++++++++++++++ 10 files changed, 93 insertions(+), 20 deletions(-) create mode 100644 code/lc25.java diff --git a/code/lc148.java b/code/lc148.java index 7ab9390..290e85e 100644 --- a/code/lc148.java +++ b/code/lc148.java @@ -27,12 +27,12 @@ public class lc148 { } ListNode slow = head; ListNode fast = head.next; - while( fast.next!=null && fast.next.next!=null ){ //把链表分成两半 + while( fast!=null && fast.next!=null ){ //把链表分成两半 slow = slow.next; fast = fast.next.next; } ListNode l2 = sortList(slow.next); - slow.next = null; + slow.next = null; //别忘了这要断开 ListNode l1 = sortList(head); return mergeList(l1, l2); } diff --git a/code/lc153.java b/code/lc153.java index 2037e08..cc9085f 100644 --- a/code/lc153.java +++ b/code/lc153.java @@ -4,8 +4,9 @@ package code; * 题意：反转数组找最小值 * 难度：Medium * 分类：Array, Binary Search - * 思路： + * 思路：想清楚，不用那么多判断 * Tips：边界条件想清楚 + * nums[mid]和nums[right]比就行了 */ public class lc153 { public int findMin(int[] nums) { @@ -13,18 +14,10 @@ public class lc153 { int right = nums.length-1; while(leftnums[right]){ + left = mid+1; } } return nums[left]; diff --git a/code/lc206.java b/code/lc206.java index 341aa8d..a21d5f9 100644 --- a/code/lc206.java +++ b/code/lc206.java @@ -6,6 +6,7 @@ package code; * 分类：Linked List * 思路：2中方法：设置一个快走一步的快指针，注意赋值操作顺序。还有一种递归的方法。 * Tips：递归的方法有点绕，多看下 + * lc25, lc206 */ public class lc206 { public class ListNode { diff --git a/code/lc236.java b/code/lc236.java index fb0b274..037e49e 100644 --- a/code/lc236.java +++ b/code/lc236.java @@ -9,6 +9,7 @@ import java.util.*; * 分类：Tree * 思路：递归，迭代两种方法 * Tips：注意递归时怎么返回。很经典的题目。 + * */ public class lc236 { public class TreeNode { @@ -18,7 +19,7 @@ public class lc236 { TreeNode(int x) { val = x; } } public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {//递归 - if( root==null || root==p || root==q ) + if( root==null || root==p || root==q ) //注意这个条件 return root; TreeNode left = lowestCommonAncestor(root.left, p, q); TreeNode right = lowestCommonAncestor(root.right, p, q); diff --git a/code/lc25.java b/code/lc25.java new file mode 100644 index 0000000..57d1321 --- /dev/null +++ b/code/lc25.java @@ -0,0 +1,39 @@ +package code; +/* + * 25. Reverse Nodes in k-Group + * 题意：每k个反转一下，不足k的不反转，直接接上 + * 难度：Hard + * 分类：Linked List + * 思路：递归调用反转，反转完下一段的返回节点，节点这一段上 + * Tips：lc25, lc206 + */ +public class lc25 { + public class ListNode { + int val; + ListNode next; + ListNode(int x) { + val = x; + } + } + + public ListNode reverseKGroup(ListNode head, int k) { + ListNode curr = head; + int count = 0; + while (curr != null && count != k) { // 找下一段要反转的起始节点 + curr = curr.next; + count++; + } + if (count == k) { // 不足k的不执行 + curr = reverseKGroup(curr, k); // 递归调用，反转下一段，并返回翻转后的头结点 + // 反转当前段 + while (count-- > 0) { // 链表反转的思路 + ListNode tmp = head.next; + head.next = curr; + curr = head; + head = tmp; + } + head = curr; + } + return head; + } +} diff --git a/code/lc300.java b/code/lc300.java index aeb09d6..1003cf4 100644 --- a/code/lc300.java +++ b/code/lc300.java @@ -14,7 +14,7 @@ public class lc300 { if(nums.length<2) return nums.length; int[] dp = new int[nums.length]; //dp[i] 存储以nums[i]结尾的最大长度 - Arrays.fill(dp,1); + Arrays.fill(dp,1); //记住fill 1 int res = 1; for (int i = 1; i < nums.length ; i++) { for (int j = 0; j < i ; j++) { diff --git a/code/lc315.java b/code/lc315.java index 6a5223d..f8a3728 100644 --- a/code/lc315.java +++ b/code/lc315.java @@ -11,7 +11,7 @@ import java.util.List; * 思路：两种思路，一种用二叉搜索树这类数据结构 https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76580/9ms-short-Java-BST-solution-get-answer-when-building-BST * 一种归并排序的思路，归并的时候统计左右交换数目。如果一个数从这个数的右边交换到左边，则+1。因为有重复数字，所以用将index进行排序 * https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76583/11ms-JAVA-solution-using-merge-sort-with-explanation - * 再有一种复杂度稍微高点的思路，从后往前插入排序，插入的时候二分搜索 + * 再有一种复杂度稍微高点的思路，从后往前插入排序，插入的时候二分搜索，插入其实是不行的，因为插入操作以后还要移动value位置，又是一个O(N)的操作 * https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76576/My-simple-AC-Java-Binary-Search-code * Tips：好难呀，我日！ */ diff --git a/code/lc72.java b/code/lc72.java index e75490a..505cc6d 100644 --- a/code/lc72.java +++ b/code/lc72.java @@ -9,7 +9,7 @@ package code; */ public class lc72 { public static void main(String[] args) { - System.out.println(minDistance("intention","execution")); + System.out.println(minDistance("horse","ros")); } public static int minDistance(String word1, String word2) { int[][] dp = new int[word1.length()+1][word2.length()+1]; diff --git a/code/lc75.java b/code/lc75.java index 31c60b2..b66cc4f 100644 --- a/code/lc75.java +++ b/code/lc75.java @@ -18,7 +18,7 @@ public class lc75 { public static void sortColors(int[] nums) { int l = 0; int r = nums.length-1; - for (int i = 0; i <= r ; i++) { // i=0){ + AdjustTree(nums, nums.length-1, pos); + pos--; + } + //排序，每次找出最大的，从堆中去掉，调整堆 + int cur = nums.length-1; + while(cur>=0){ + //交换位置 + int temp = nums[0]; + nums[0] = nums[cur]; + nums[cur] = temp; + cur--; + AdjustTree(nums, cur, 0); + } + return nums; + } + + public void AdjustTree(int[] nums, int len, int pos){ //调整堆 + int pos_exchange = pos*2+1; + while(pos_exchange<=len){ //left + if(pos_exchange+1<=len&&nums[pos_exchange+1]>nums[pos_exchange]){ //比较左右节点，挑出来大的 + pos_exchange += 1; + } + if(nums[pos_exchange]>nums[pos]){ //和父节点比较 + int temp = nums[pos]; + nums[pos] = nums[pos_exchange]; + nums[pos_exchange] = temp; + pos = pos_exchange; + pos_exchange = pos*2+1; + }else{ + break; + } + } + } } -- GitLab