Skip to content

Navigation Menu

Sign in
Appearance settings

Search code, repositories, users, issues, pull requests...

Provide feedback

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly
Sign up
Appearance settings

Commit cfc97c8

Browse files
20200816
1 parent 1cf255d commit cfc97c8

File tree

3 files changed

+104
-0
lines changed

3 files changed

+104
-0
lines changed
Lines changed: 50 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,50 @@
1+
/*
2+
* @lc app=leetcode id=1145 lang=java
3+
*
4+
* [1145] Binary Tree Coloring Game
5+
*/
6+
7+
// @lc code=start
8+
/**
9+
* Definition for a binary tree node.
10+
* public class TreeNode {
11+
* int val;
12+
* TreeNode left;
13+
* TreeNode right;
14+
* TreeNode() {}
15+
* TreeNode(int val) { this.val = val; }
16+
* TreeNode(int val, TreeNode left, TreeNode right) {
17+
* this.val = val;
18+
* this.left = left;
19+
* this.right = right;
20+
* }
21+
* }
22+
*/
23+
class Solution {
24+
/*
25+
如果一个节点确定之后,其subtree对方都不能访问
26+
将left, right, parent看作三个subtree
27+
比较是否有一边subtree的数量大于n/2
28+
if countLeft or countRight are bigger than n/2, player 2 chooses this child of the node and will win.
29+
If countLeft + countRight + 1 is smaller than n/2, player 2 chooses the parent of the node and will win;
30+
otherwise, lose
31+
*/
32+
int l, r;
33+
public boolean btreeGameWinningMove(TreeNode root, int n, int x) {
34+
count(root, x);
35+
return Math.max(l, Math.max(r, n - l - r - 1)) > n / 2;
36+
}
37+
38+
private int count(TreeNode root, int target) {
39+
if (root == null) return 0;
40+
int left = count(root.left, target);
41+
int right = count(root.right, target);
42+
if (root.val == target) {
43+
l = left;
44+
r = right;
45+
}
46+
return left + right + 1;
47+
}
48+
}
49+
// @lc code=end
50+
Lines changed: 29 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,29 @@
1+
/*
2+
* @lc app=leetcode id=123 lang=java
3+
*
4+
* [123] Best Time to Buy and Sell Stock III
5+
*/
6+
7+
// @lc code=start
8+
class Solution {
9+
/*
10+
假设手上最开始只有0元钱,那么如果买入股票的价格为price,手上的钱需要减去这个price,如果卖出股票的价格为price,
11+
手上的钱需要加上这个price。然后不断更新第一次买,第一次卖,第二次买,第二次卖的钱。
12+
time: O(n)
13+
space: O(1)
14+
*/
15+
public int maxProfit(int[] prices) {
16+
if (prices == null || prices.length == 0) return 0;
17+
int buy1= Integer.MIN_VALUE, buy2= Integer.MIN_VALUE;
18+
int sell1 = 0, sell2 = 0;
19+
for (int price : prices) {
20+
buy1 = Math.max(buy1, -price);
21+
sell1 = Math.max(sell1, buy1 + price);
22+
buy2 = Math.max(buy2, sell1 - price);
23+
sell2 = Math.max(sell2, buy2 + price);
24+
}
25+
return sell2;
26+
}
27+
}
28+
// @lc code=end
29+
Lines changed: 25 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,25 @@
1+
/*
2+
* @lc app=leetcode id=1347 lang=java
3+
*
4+
* [1347] Minimum Number of Steps to Make Two Strings Anagram
5+
*/
6+
7+
// @lc code=start
8+
class Solution {
9+
public int minSteps(String s, String t) {
10+
int[] count = new int[26];
11+
for (int i = 0; i < s.length(); ++i) {
12+
count[s.charAt(i) - 'a']++;
13+
count[t.charAt(i) - 'a']--;
14+
}
15+
int res = 0;
16+
for (int c : count) {
17+
if (c > 0) {
18+
res += c;
19+
}
20+
}
21+
return res;
22+
}
23+
}
24+
// @lc code=end
25+

0 commit comments

Comments
(0)

AltStyle によって変換されたページ (->オリジナル) /