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Commit 4ca0257

Browse files
committed
20190715
1 parent abfdac6 commit 4ca0257

19 files changed

+193
-46
lines changed

‎code/lc102.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -23,7 +23,7 @@ public List<List<Integer>> levelOrder(TreeNode root) {
2323
if(root==null)
2424
return res;
2525
qu.add(root);
26-
while(!qu.isEmpty()){
26+
while(!qu.isEmpty()){//两个while
2727
int size = qu.size();
2828
List<Integer> temp = new ArrayList<>();
2929
while(size>0){

‎code/lc113.java

Lines changed: 11 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -20,20 +20,20 @@ public class TreeNode {
2020
}
2121
}
2222
public List<List<Integer>> pathSum(TreeNode root, int sum) {
23-
List<List<Integer>> res = new ArrayList<>();
24-
helper(res, root, sum, 0, new ArrayList<>());
23+
List<List<Integer>> res = new ArrayList();
24+
helper(res, new ArrayList(), root, sum);
2525
return res;
2626
}
27-
public void helper(List<List<Integer>> res, TreeNoderoot, intsum, int curr, List<Integer> curr_ls) {
27+
public void helper(List<List<Integer>> res, List<Integer> cur, TreeNoderoot, int sum){
2828
if(root==null) return;
29-
curr_ls.add(root.val);
30-
if(curr+root.val==sum && root.left==null && root.right==null) { //到叶子节点
31-
res.add(new ArrayList<>(curr_ls));
32-
curr_ls.remove(curr_ls.size()-1);
33-
return;
29+
cur.add(root.val);
30+
if(root.left==null&&root.right==null&&root.val==sum){ //到叶子节点
31+
res.add(new ArrayList(cur));
32+
}else{
33+
helper(res, cur, root.left, sum-root.val);
34+
helper(res, cur, root.right, sum-root.val);
3435
}
35-
helper(res, root.left, sum, curr+root.val, curr_ls);
36-
helper(res, root.right, sum, curr+root.val, curr_ls);
37-
curr_ls.remove(curr_ls.size()-1);
36+
cur.remove(cur.size()-1);
37+
return;
3838
}
3939
}

‎code/lc123.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -1,6 +1,6 @@
11
package code;
22
/*
3-
* 122. Best Time to Buy and Sell Stock III
3+
* 123. Best Time to Buy and Sell Stock III
44
* 题意:买卖股票最大利润,只能买卖2次
55
* 难度:Hard
66
* 分类:Array, Dynamic Programming

‎code/lc131.java

Lines changed: 1 addition & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -10,6 +10,7 @@
1010
* 思路:典型回溯法,注意向res添加内容时要重新new一下
1111
* Tips: lc5, lc9, lc125, lc131, lc234, lc647
1212
* lc39
13+
* lc132
1314
* 判断是否为回文的方法:
1415
* 1. 从中心往两边扩充,中心可能是一个字符,也可能是两个字符
1516
* 2. dp,利用之前计算的结果,只判断边缘两个字符是否相等,从后往前dp

‎code/lc132.java

Lines changed: 52 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,52 @@
1+
package code;
2+
import java.util.Arrays;
3+
4+
/*
5+
* 132. Palindrome Partitioning II
6+
* 题意:最少切几刀,回文串
7+
* 难度:Hard
8+
* 分类:Dynamic Programming
9+
* 思路:和lc300思路很像,只是多了判断。dp[i]表示从0到i这一段切法的最小值,每次遍历前边的结果,看能否接上。
10+
* 可以把判断回文和dp两个合并,因为用dp的形式判断回文
11+
* Tips:lc300
12+
* bingo!
13+
*/
14+
15+
public class lc132 {
16+
public int minCut(String s) {
17+
int[] dp = new int[s.length()+1];
18+
Arrays.fill(dp, s.length()); //fill最大值
19+
dp[0] = 0;
20+
for(int i=1; i<dp.length; i++){
21+
for(int j=0; j<=i; j++){ //注意<=
22+
if(dp[j]!=s.length() && isPalindrome(s.substring(j,i)))
23+
dp[i] = Math.min(dp[i], dp[j]+1);
24+
}
25+
}
26+
return dp[s.length()]-1; //要-1 最后一刀是在末尾
27+
}
28+
29+
public Boolean isPalindrome(String str){
30+
if(new StringBuilder(str).reverse().toString().equals(str)) return true; //reverse一下
31+
else return false;
32+
}
33+
34+
public int minCut2(String s) {
35+
char[] c = s.toCharArray();
36+
int n = c.length;
37+
int[] cut = new int[n];
38+
boolean[][] pal = new boolean[n][n];
39+
40+
for(int i = 0; i < n; i++) {
41+
int min = i;
42+
for(int j = 0; j <= i; j++) {
43+
if(c[j] == c[i] && (j + 1 > i - 1 || pal[j + 1][i - 1])) {
44+
pal[j][i] = true;
45+
min = j == 0 ? 0 : Math.min(min, cut[j - 1] + 1);
46+
}
47+
}
48+
cut[i] = min;
49+
}
50+
return cut[n - 1];
51+
}
52+
}

‎code/lc139.java

Lines changed: 14 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -6,8 +6,10 @@
66
* 分类:Dynamic Programming
77
* 思路:动态规划
88
* Tips:巧妙的方法,防止了复杂的操作,通过遍历之前计算出来的结果
9+
* 递归的方法本质和dp是一样的,记住用备忘录算法,把之前的结果记下来
910
* lc140
1011
*/
12+
import java.util.HashMap;
1113
import java.util.List;
1214

1315
public class lc139 {
@@ -22,4 +24,16 @@ public boolean wordBreak(String s, List<String> wordDict) {
2224
}
2325
return dp[s.length()];
2426
}
27+
28+
HashMap<String, Boolean> hm = new HashMap();
29+
public boolean wordBreak2(String s, List<String> wordDict) {
30+
if(hm.containsKey(s)) return hm.get(s);
31+
if(s.length() == 0) return true;
32+
Boolean flag = false;
33+
for(String word:wordDict){
34+
if(s.startsWith(word)) flag = flag||wordBreak(s.substring(word.length()), wordDict); //注意函数 startsWith
35+
}
36+
hm.put(s, flag);
37+
return flag;
38+
}
2539
}

‎code/lc141.java

Lines changed: 5 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -14,15 +14,13 @@ public class ListNode {
1414
ListNode(int x) { val = x; }
1515
}
1616
public boolean hasCycle(ListNode head) {
17-
if(head==null)
18-
return false;
19-
ListNode faster = head;
17+
if(head==null||head.next==null) return false;
2018
ListNode slow = head;
21-
while( faster.next!=null && faster.next.next!=null){ //注意判断条件,slow一定不等于null,不用判断了
19+
ListNode fast = head.next;
20+
while(fast!=null&&fast.next!=null){ //注意判断条件,slow一定不等于null,不用判断了
2221
slow = slow.next;
23-
faster = faster.next.next;
24-
if(slow==faster)
25-
return true;
22+
fast = fast.next.next;
23+
if(fast==slow) return true;
2624
}
2725
return false;
2826
}

‎code/lc142.java

Lines changed: 19 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -32,4 +32,23 @@ public ListNode detectCycle(ListNode head) {
3232
}
3333
return null;
3434
}
35+
36+
public ListNode detectCycle2(ListNode head) {
37+
if(head==null||head.next==null) return null;
38+
ListNode slow = head;
39+
ListNode fast = head.next;
40+
while(fast!=null&&fast.next!=null){
41+
slow = slow.next;
42+
fast = fast.next.next;
43+
if(slow==fast){
44+
slow = slow.next;
45+
while(head!=slow){
46+
head = head.next;
47+
slow = slow.next;
48+
}
49+
return slow;
50+
}
51+
}
52+
return null;
53+
}
3554
}

‎code/lc148.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -25,7 +25,7 @@ public ListNode sortList(ListNode head) {
2525
if( head==null || head.next == null ){
2626
return head;
2727
}
28-
ListNode slow = head;
28+
ListNode slow = head;//记一下
2929
ListNode fast = head.next;
3030
while( fast!=null && fast.next!=null ){ //把链表分成两半
3131
slow = slow.next;

‎code/lc207.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -23,7 +23,7 @@ public static boolean canFinish(int numCourses, int[][] prerequisites) {
2323
for (int i = 0; i < prerequisites.length ; i++) {
2424
int node1 = prerequisites[i][0];
2525
int node2 = prerequisites[i][1];
26-
graph[node2][node1] = 1;
26+
graph[node2][node1] = 1;//存下邻接矩阵,方便后续查找是否有边
2727
indegree[node1]++; //存下入度,入度为0时,表示该课程可以修
2828
}
2929
Stack<Integer> st = new Stack();

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