LeetCode Longest Word in Dictionary through Deleting Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographical order. If there is no possible result, return the empty string. Example 1:

Input:
s = "abpcplea", d = ["ale","apple","monkey","plea"]
Output:
"apple"

Input:
s = "abpcplea", d = ["a","b","c"]
Output:
"a"

1. All the strings in the input will only contain lower-case letters.
2. The size of the dictionary won’t exceed 1,000.
3. The length of all the strings in the input won’t exceed 1,000.

289. Game of Life 289. Game of Life

According to the Wikipedia’s article: “The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970.”

Given a board with m by n cells, each cell has an initial state live (1) or dead (0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

1. Any live cell with fewer than two live neighbors dies, as if caused by under-population.
2. Any live cell with two or three live neighbors lives on to the next generation.
3. Any live cell with more than three live neighbors dies, as if by over-population..
4. Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

Write a function to compute the next state (after one update) of the board given its current state. The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously.

Example:

Input: 
[
  [0,1,0],
  [0,0,1],
  [1,1,1],
  [0,0,0]
]
Output: 
[
  [0,0,0],
  [1,0,1],
  [0,1,1],
  [0,1,0]
]

Follow up:

1. Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.
2. In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches the border of the array. How would you address these problems? 289. Game of Life

本题有关康威生命游戏，第一次听说。。。 给一个二维矩阵，0表示死亡，1表示活着。每个细胞根据周围8个细胞的状态，有如下跳转规则：

1. 活着的细胞周围有少于2个活着的细胞，则该细胞死亡
2. 活着的细胞周围有2个或3个活着的细胞，则该细胞继续活着
3. 活着的细胞周围有多于3个活着的细胞，则该细胞死亡
4. 死亡的细胞周围有3个活着的细胞，则该细胞复活

最简单的方法是复制一个矩阵，遍历原矩阵然后更新新矩阵。但本题的难点是，要在in-place的情况下更新原矩阵。如果还是像之前那样更新的话，当前细胞的状态更新之后，会影响到后续细胞的判断，不可行。 参考网上题解，使用编码方法。 因为输入是以int来存储细胞状态的，int可以存储的状态数就很多了，不仅仅局限于0和1。 假设细胞状态重新编码如下：

• 0：0→0，原来死亡，现在还死亡
• 1：1→1，原来活着，现在还活着
• 2：1→0，原来活着，现在死亡
• 3：0→1，原来死亡，现在活着

则还是像之前一样，遍历细胞的周围8个点，根据以上编码可以知道周围细胞更新前的状态，比如某个邻居的状态是2，则知道该邻居更新前是1活着的。 所有细胞都更新完之后，需要复原成0，1编码，则状态为0和2表示新状态是死亡0，状态是1和3表示新状态是活着1。这可以用新状态对2取余数得到。 代码如下：

class Solution {
public:
    void gameOfLife(vector<vector<int> >& board)
    {
        int m = board.size(), n = 0;
        if (m != 0)
            n = board[0].size();
        if (m == 0 || n == 0)
            return;
        vector<vector<int> > dirs = { { -1, -1 }, { -1, 0 }, { -1, 1 }, { 0, -1 }, { 0, 1 }, { 1, -1 }, { 1, 0 }, { 1, 1 } };
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                int lives = 0;
                for (int k = 0; k < dirs.size(); ++k) {
                    int x = i + dirs[k][0], y = j + dirs[k][1];
                    if (x >= 0 && x < m && y >= 0 && y < n && (board[x][y] == 1 || board[x][y] == 2))
                        ++lives;
                }
                if (board[i][j] == 1 || board[i][j] == 2) {
                    if (lives < 2 || lives > 3)
                        board[i][j] = 2;
                    else
                        board[i][j] = 1;
                }
                else if (lives == 3)
                    board[i][j] = 3;
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                board[i][j] %= 2;
            }
        }
    }
};

本代码提交AC，用时3MS。

二刷。 很简单，也不用定什么规则，直接按位编码，第0位是原来的状态，0为死1为生，第1位是更新后的状态，依然是0为死1为生，所有状态更新完之后，右移一位即可。代码如下：

class Solution {
public:
	void gameOfLife(vector<vector<int>>& board) {
		int m = board.size(), n = board[0].size();

		for (int i = 0; i < m; ++i) {
			for (int j = 0; j < n; ++j) {

				int lives = 0;
				int u = max(0, i - 1), v = max(0, j - 1);
				int x = min(m, i - 1 + 3), y = min(n, j - 1 + 3);
				for (int a = u; a < x; ++a) {
					for (int b = v; b < y; ++b) {
						if (a == i && b == j)continue;
						if ((board[a][b] & 1) == 1)++lives;
					}
				}

				int curlive = (board[i][j] & 1);
				if (curlive == 1 && lives < 2) {
					board[i][j] = (board[i][j] | (0 << 1));
				}
				else if (curlive == 1 && (lives == 2 || lives == 3)) {
					board[i][j] = (board[i][j] | (1 << 1));
				}
				else if (curlive == 1 && (lives > 3)) {
					board[i][j] = (board[i][j] | (0 << 1));
				}
				else if (curlive == 0 && lives == 3) {
					board[i][j] = (board[i][j] | (1 << 1));
				}

			}
		}

		for (int i = 0; i < m; ++i) {
			for (int j = 0; j < n; ++j) {
				board[i][j] >>= 1;
			}
		}

	}
};

本代码提交AC，用时0MS。

LeetCode Non-overlapping Intervals Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping. Note:

1. You may assume the interval’s end point is always bigger than its start point.
2. Intervals like [1,2] and [2,3] have borders “touching” but they don’t overlap each other.

Input: [ [1,2], [2,3], [3,4], [1,3] ]
Output: 1
Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

Input: [ [1,2], [1,2], [1,2] ]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.

Input: [ [1,2], [2,3] ]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

LeetCode Find Right Interval Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the “right” of i. For any interval i, you need to store the minimum interval j’s index, which means that the interval j has the minimum start point to build the “right” relationship for interval i. If the interval j doesn’t exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array. Note:

1. You may assume the interval’s end point is always bigger than its start point.
2. You may assume none of these intervals have the same start point.

Input: [ [1,2] ]
Output: [-1]
Explanation: There is only one interval in the collection, so it outputs -1.

Input: [ [3,4], [2,3], [1,2] ]
Output: [-1, 0, 1]
Explanation: There is no satisfied "right" interval for [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point;
For [1,2], the interval [2,3] has minimum-"right" start point.

Input: [ [1,4], [2,3], [3,4] ]
Output: [-1, 2, -1]
Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
For [2,3], the interval [3,4] has minimum-"right" start point.

LeetCode Lexicographical Numbers Given an integer n, return 1 – n in lexicographical order. For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9]. Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000.

LeetCode Bulb Switcher There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it’s off or turning off if it’s on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds. Example:

Given n = 3.
At first, the three bulbs are [off, off, off].
After first round, the three bulbs are [on, on, on].
After second round, the three bulbs are [on, off, on].
After third round, the three bulbs are [on, off, off].
So you should return 1, because there is only one bulb is on.