1. Sum Root to Leaf Numbers
Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
Note: A leaf is a node with no children.
Example:
Input: [1,2,3]
1
/ \
2 3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
Example 2:
Input: [4,9,0,5,1]
4
/ \
9 0
/ \
5 1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.
思路：本来想用递归解决这个问题，结果在传参时出现了一个很典型的错误，结果还是点来了discussion，瞬间清朗很多，果然大佬就是大佬。
public int sumNumbers(TreeNode root) {
return sum(root, 0);
}
public int sum(TreeNode n, int s){
if (n == null) return 0;
if (n.right == null && n.left == null) return s*10 + n.val; // 叶节点的情形
return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val); // 不是叶节点的情形
}
感觉对于树有关的题目，很多都可以用递归来解决，因为在处理完了根节点后，对于它的左子节点和右子节点的处理流程和根节点差不多，关键是着其中参数的变化。
2. Surrounded Regions
Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.
A region is captured by flipping all 'O's into 'X's in that surrounded region.
Example:
X X X X X O O X X X O X X O X X
After running your function, the board should be:
X X X X X X X X X X X X X O X X
思路：类似于围棋一样，基本思路是先将整个矩阵中没有围住的 O (也就是临接边界的) 置为 * ，再将被围住的(其余的) O 置为 X ，再将 * 置回 O 即可，需要注意的是在第一步寻找没有被围住的 O 时，需要用到DFS或者BFS来搜索。
public void solve(char[][] board) {
if (board.length == 0 || board[0].length == 0)
return;
if (board.length < 2 || board[0].length < 2)
return;
int m = board.length, n = board[0].length;
//Any 'O' connected to a boundary can't be turned to 'X', so ...
//Start from first and last column, turn 'O' to '*'.
for (int i = 0; i < m; i++) {
if (board[i][0] == 'O')
boundaryDFS(board, i, 0);
if (board[i][n-1] == 'O')
boundaryDFS(board, i, n-1);
}
//Start from first and last row, turn '0' to '*'
for (int j = 0; j < n; j++) {
if (board[0][j] == 'O')
boundaryDFS(board, 0, j);
if (board[m-1][j] == 'O')
boundaryDFS(board, m-1, j);
}
//post-prcessing, turn 'O' to 'X', '*' back to 'O', keep 'X' intact.
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] == 'O')
board[i][j] = 'X';
else if (board[i][j] == '*')
board[i][j] = 'O';
}
}
}
//Use DFS algo to turn internal however boundary-connected 'O' to '*';
private void boundaryDFS(char[][] board, int i, int j) {
if (i < 0 || i > board.length - 1 || j <0 || j > board[0].length - 1)
return;
if (board[i][j] == 'O') // 感觉这个判断条件可以省略
board[i][j] = '*';
if (i > 1 && board[i-1][j] == 'O') // 向四个相邻的方向DFS，DFS是递归，BFS是队列。上
boundaryDFS(board, i-1, j);
if (i < board.length - 2 && board[i+1][j] == 'O') // 下
boundaryDFS(board, i+1, j);
if (j > 1 && board[i][j-1] == 'O') // 左
boundaryDFS(board, i, j-1);
if (j < board[i].length - 2 && board[i][j+1] == 'O' ) // 右
boundaryDFS(board, i, j+1);
}
注意的是上面往四个方向DFS时的if判断条件，这里用 length-2 而不是 length-1 是为了避免过深的DFS而导致的stackOverflow。
3. Single Number II
Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,3,2] Output: 3
Example 2:
Input: [0,1,0,1,0,1,99] Output: 99
思路：本来为是比较简单的一题，结果还是各种恪手，算法这玩意儿，真的是好难学(┬＿┬)。看了一下discuss，主要是利用位运算符，既然题目要求在线性时间内完成，即扫描一遍整个数组就能得出结果。
public int singleNumber(int[] nums) {
int ans = 0;
for(int i = 0; i < 32; i++) { // 从低位到高位处理
int sum = 0;
for(int j = 0; j < nums.length; j++) {
if(((nums[j] >> i) & 1) == 1) {
sum++;
sum %= 3; // 这里可以扩展
}
}
if(sum != 0) {
ans |= sum << i; // 确定每个位置最终是0还是1后，还原值
}
}
return ans;
}
选择了discuss中最容易理解的一个，主要思想就是将整数转化成32位的二进制来考虑，因为数组中的元素要么出现了3次要么出现了一次，所以任何一位上的数字(0或者1)出现的总次数是3的倍数或者3的倍数加1，所以统计总共每个位置上出现的1的个数即可，如果到了次数3就置为0，这样统计下来，那些出现次数为3的数每个二进制位置上都是0，只要出现一次的数字的二进制位留在了最后处理结果上。
