Pocket Gems专题

 

// Maximum Product Subarray 
public int maxProduct(int[] nums) {
    int result = nums[0];
    int maxHere = nums[0];
    int minHere = nums[0];
    for(int i=1; i<nums.length; i++) {
        int a = nums[i] * maxHere;
        int b = nums[i] * minHere;
        maxHere = Math.max(nums[i], Math.max(a, b));
        minHere = Math.min(nums[i], Math.min(a, b));
        result = Math.max(result, maxHere);
    }
    return result;
}

// Neighboring classroom, given a map of m * m, and n classrooms, 
//determine if every classroom belongs to one single component; 
//bounds: (1)classroom is at most 3 * 3 in area 
//(2) no overlapping classrooms
// (3) classrooms are at least 5% of m * m in total area 
//(4) isConnected returns true only when two classroom shares a common EDGE.
public static class Room {
	int x, y, h, w;
}
private static int[][] dir = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public boolean isConnected(int m, int n, List<Room> rooms) {
	boolean[][] board = new boolean[m][n];
	int area = 0;
	for(Room room : rooms) {
		for(int i=room.y-room.h+1; i<=room.y; i++) {
			for(int j=room.x; j<=room.x+room.w-1; j++) {
				board[i][j] = true;
			}
		}
		area += room.h * room.w;
	}
	
	Queue<Integer> queue = new LinkedList<>();
	int start = rooms.get(0).y * n + rooms.get(0).x;
	queue.offer(start);
	int sum = 0;
	
	while(!queue.isEmpty()) {
		sum++;
		int point = queue.poll();
		int y = point / n;
		int x = point % n;
		board[y][x] = false;
		for(int i=0; i<dir.length; i++) {
			int row = y + dir[i][0];
			int col = x + dir[i][1];
			if(row >= 0 && row < m && col >=0 && col < n && board[row][col]) {
				queue.offer(row*n+col);
			}
		}
	}
	
	return sum == area;
}

// sliding windows max
public static int[] slidingMax(int[] A, int w) {
	int n = A.length;
	w = Math.min(n, w);
	int k = n - w + 1;
	int[] max = new int[k];
	Deque<Integer> deq = new ArrayDeque<>();
	for(int i=0; i<n; i++) {
		while(!deq.isEmpty() && A[deq.getLast()] <= A[i]) { // A[deq.getLast()] >= A[i] for slidingMin
			deq.removeLast();
		}
		deq.addLast(i);
		if(i < w-1) continue;
		while(!deq.isEmpty() && i-w>=deq.getFirst()) {
			deq.removeFirst();
		}
		max[i-w+1] = A[deq.getFirst()];
	}
	return max;
}

// given an array, return the starting and ending index of all subarrays that sums to 0;
public void getZeroSumIndex(int[] A) {
	int n = A.length;
	int[] sum = new int[n+1];
	Map<Integer, List<Integer>> map = new HashMap<>();
	Set<Integer> result = new HashSet<>();
	for(int i=0; i<n; i++) {
		sum[i+1] = A[i] + sum[i];
		if(sum[i+1] == 0) {
			result.add(0*31 +i);
		}
		if(A[i] == 0) {
			result.add(i*31 + i);
		}
		List<Integer> list = map.get(sum[i+1]);
		if(list == null) {
			list = new ArrayList<>();
			map.put(sum[i+1], list);
		} else {
			for(int index : list) {
				result.add((index+1)*31 + i);
			}
		}
		list.add(i);
	}
	
	for(int num: result) {
		System.out.println(num/31 + ", " + num%31);
	}
}

// word break 1
// 重要的是要写出时间复杂度 递归（2^n）? 和worst case(如aaac, 字典是（"a", "aa", "aaa"）)
// Time: O(n^2)
public boolean wordBreak(String s, Set<String> wordDict) {
    int n = s.length();
    boolean[] f = new boolean[n+1];
    f[0] = true;
    for(int i=1; i<=n; i++) {
        for(int j=0; j<i; j++) {
            String word = s.substring(j, i);
            f[i] = f[j] && wordDict.contains(word);
            if(f[i]) break;
        }
    }
    return f[n];
}

public static boolean wordBreak(String s, Set<String> dict){
	//Base case
	if(dict.contains(s)) return true;
	for(int i = 0; i < s.length(); i++){
		String sstr = s.substring(0, i);
		if(dict.contains(sstr) && wordBreak(s.substring(i), dict))
			return true;
	}
	return false;
}

// word break 1的另外一种写法 O(M*N), 
// Time: O(string length * dict size)
public boolean wordBreak(String s, Set<String> dict) {
        boolean[] t = new boolean[s.length()+1];
        t[0] = true; //set first to be true, why?
        //Because we need initial state
        for(int i=0; i<s.length(); i++){
            //should continue from match position
            if(!t[i]) continue;
            for(String a: dict){
                int len = a.length();
                int end = i + len;
                if(end > s.length()) continue;
                if(t[end]) continue;
                if(s.substring(i, end).equals(a)){
                    t[end] = true;
                }
            }
        }
        return t[s.length()];
    }
}

// word break II recursive method, Time is O(n^2) 
public List<String> wordBreak(String s, Set<String> dict) {
    List<String> list = new ArrayList<>();
    boolean[] f = new boolean[s.length()+1];
    Arrays.fill(f, true);
    breakWord(list, dict, f, s, 0, "");
    return list;
}

public void breakWord(List<String> list, Set<String> dict, boolean[] f, String s, int start, String t) {
    if(start == s.length()) {
        list.add(t.substring(1));
        return;
    }
    for(int i=start+1; i<=s.length(); i++) {
        String word = s.substring(start, i);
        if(dict.contains(word) && f[i]) {
            int size = list.size();
            breakWord(list, dict, f, s, i, t+" "+word);
            if(list.size() == size) f[i] = false;
        }
    }
}

// longest palindrome substring
public String longestPalindrome(String s) {
    String res = "";
    for(int i=0; i<s.length(); i++) {
        String str = palindromeAtCenter(s, i, i);
        if(str.length() > res.length()) {
            res = str;
        }
        
        str = palindromeAtCenter(s, i, i+1);
        if(str.length() > res.length()) {
            res = str;
        }
    }
    return res;
}

private String palindromeAtCenter(String s, int c1, int c2) {
    while(c1>=0 && c2<s.length() && s.charAt(c1) == s.charAt(c2)) {
        c1--;
        c2++;
    }
    return s.substring(c1+1, c2);
}

// reservior sampling
public int[] samplingK(Scanner s, int k) {
	int[] res = new int[k];
	int i = 0;
	while (i < k) {
		res[i++] = s.nextInt();
	}
	Random r = new Random();
	while(s.hasNext()) {
		int num = s.nextInt();
		int rand = r.nextInt(i+1); // important: inclusive range 
		if(rand < k) {
			res[rand] = num;
		}
	}
	return res;
}

// topological sorting
public String getOrderedString(String[] strs, int charSize) {
	DirectedGraph g = new DirectedGraph(charSize);
	boolean[] visited = new boolean[charSize];
	Arrays.fill(visited, true); //~~
	for(String s:strs) {
		if(s.isEmpty()) continue;
		visited[s.charAt(0)-'a'] = false; //~~
		for(int i=1; i<s.length(); i++) {
			visited[s.charAt(i)-'a'] = false; //~~
			g.addEdge(s.charAt(i-1)-'a', s.charAt(i)-'a');
		}
	}
	Stack<Integer> stack = new Stack<>();
	for(int i=0; i<charSize; i++) {
		if(!visited[i])
			toposort(g, i, visited, stack);
	}
	
	StringBuilder sb = new StringBuilder();
	while(!stack.isEmpty()) {
		sb.append((char)(stack.pop()+ 'a'));
	}
	System.out.println(sb.toString());
	return sb.toString();
}

public void toposort(DirectedGraph g, int v, boolean[] visited, Stack<Integer> stack) {
	visited[v] = true;
	for(int u : g.adj[v]) {
		if(!visited[u]) {
			toposort(g, u, visited, stack);
		}
	}
	stack.push(v);
}

// room isConnected
// 第四轮的题还问到了时间复杂度
// 对于访问到的每个点判断是否满足小于k需要O(k)，
// 所有访问到的点形成的图形最长半径为n的话则有O(n2)个点，所以总共是O(n2*k)。
public class MonkeyProblem {
	static class Point {
		int x, y;
		Point(int x, int y) {
			this.x = x;
			this.y = y;
		}
		@Override
		public boolean equals(Object o) {
			if (this == o) return true;
			if (!(o instanceof Point)) return false;
			Point pair = (Point) o;
			return x == pair.x && y == pair.y;
		}
		@Override
		public int hashCode() {
			return 31 * x + y;
		}
	}
	
	public static int digitSum(int n) {
		if(n < 0) n = -n;
		int sum = 0;
		while(n != 0) {
			sum += n % 10;
			n /= 10;
		}
		return sum;
	}

	private static int[][] dir = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
	public static int countSteps(int k) {
		Set<Point> set = new HashSet<>();
		Queue<Point> queue = new LinkedList<>();
		queue.offer(new Point(0, 0));
		while(!queue.isEmpty()) {
			Point p = queue.poll();
			if(set.contains(p) || (digitSum(p.x) + digitSum(p.y)) > k) continue;
			set.add(p);
			for(int i=0; i<4; i++) {
				queue.offer(new Point(p.x+dir[i][0], p.y+dir[i][1]));
			}
		}
		return set.size();
	}

	public static void main(String[] args) {
		System.out.println(countSteps(19));
	}
}

// serialize binary tree
public String serialize(TreeNode root){  
    StringBuilder sb = new StringBuilder();  
    serialize(root, sb);  
    return sb.toString();  
}  
   
private void serialize(TreeNode x, StringBuilder sb){  
    if (x == null) {  
        sb.append("# ");  
    } else {  
        sb.append(x.val + " ");  
        serialzie(x.left, sb);  
        serialzie(x.right, sb);  
    }  
}  
   
public TreeNode deserialize(String s){  
    if (s == null || s.length() == 0) return null;  
    StringTokenizer st = new StringTokenizer(s, " ");  
    return deserialize(st);  
}  
   
private TreeNode deserialize(StringTokenizer st){  
    if (!st.hasMoreTokens())  
        return null;  
    String val = st.nextToken();  
    if (val.equals("#"))  
        return null;  
    TreeNode root = new TreeNode(Integer.parseInt(val));  
    root.left = deserialize(st);  
    root.right = deserialize(st);  
    return root;  
}  

// a?b:c  tenary tree
public TreeNode convertToTree (char[] values) {
    TreeNode root = new TreeNode(values[0]);
    TreeNode n = root;
    Stack<TreeNode> stack =  new Stack<TreeNode>();
    for (int i = 1; i < values.length; i += 2) {
        if (values[i] == '?') {
            n.left = new TreeNode (values[i + 1]);
            stack.push(n);
            n = n.left;

        }
        else if (values[i] == ':') {
            n = stack.pop();
            while (n.right != null) {
                n = stack.pop();
            }             
            n.right = new TreeNode (values[i + 1]);
            stack.push(n);
            n = n.right;
        }
    }
    return root;
}	

// mutable string
public char charAt(int index) {
    if ((index < 0) || (index >= count)) {
        throw new StringIndexOutOfBoundsException(index);
    }
    return value[index + offset];
}

public String substring(int beginIndex, int endIndex) {
	if (beginIndex < 0) {
	    throw new StringIndexOutOfBoundsException(beginIndex);
	}
	if (endIndex > count) {
	    throw new StringIndexOutOfBoundsException(endIndex);
	}
	if (beginIndex > endIndex) {
	    throw new StringIndexOutOfBoundsException(endIndex - beginIndex);
	}
	return ((beginIndex == 0) && (endIndex == count)) ? this :
	    new String(offset + beginIndex, endIndex - beginIndex, value);
}

 

 

评论

2 楼 yuanhsh 2015-06-02  

popu_12 写道

Can you please explain neighboring classroom question with example I cannot understand solution because the question is not understood? An example will help

就是房子连通问题，看所有房子能否连接成一个连通区域。
假设1代表房子所占用的区域，0代表没被房子占用的区域，那么下图的两个房子不是连通的。
00110
00110
11000
11000
但是如果下图的两个房子是连通的（每个房子都是正规的矩形）。
00110
00110
11110
11000

1 楼 popu_12 2015-06-01  

Can you please explain neighboring classroom question with example I cannot understand solution because the question is not understood? An example will help