LeetCode编程练习（2） | 程序员的自我修养 | 关注Java、大数据、机器学习

Max Points on a Line

题目

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

答案

思路：时间复杂度O(n^3)的情况下是肯定可以算出结果的，那么意味着通过空间换时间应该是可以让时间复杂度达到O(n^2)，~~这是我自己发明的定律~~。首先定义静态内部类Line，用于表示两点计算出来的直线，考虑到Double类型的精度损失问题，Line采用三个属性来保证精度上的完整性。然后重写hashCode()和equals(..)方法，使得同一条直线相等，且在Hash值上相等。如此通过2个 $n(n-1)\over 2$ 次循环即可获取到结果，时间复杂度为O(n^2)。

static class Line {
	int k;
	int b;
	int denominator;

	Line(int k, int b, int denominator) {
		this.k = k;
		this.b = b;
		this.denominator = denominator;
	}

	@Override
	public boolean equals(Object o) {
		if (this == o) return true;
		if (o == null || getClass() != o.getClass()) return false;
		Line line = (Line) o;
		if(b*line.denominator != line.b*denominator) return false;
		if(k*line.denominator != line.k*denominator) return false;
		return true;
	}

	@Override
	public int hashCode() {
		int t = denominator;
		if(t == 0) t = 1;
		int result = k/t;
		result = 31 * result + b/t;
		return result;
	}
}

public static int maxPoints(Point[] points) {
	if(points == null) return 0;
	if(points.length == 1 || points.length == 2) return points.length;

	Map<Line, Set<Point>> map = new HashMap<>();
	for(int i = 0; i < points.length; i++) {
		for(int j = i + 1; j < points.length; j++) {
			Point m = points[i];
			Point n = points[j];
			int k = n.y - m.y;
			int b = n.x * m.y - m.x * n.y;
			int denominator = n.x - m.x;
			if(denominator == 0) {
				k = Integer.MAX_VALUE;
				b = n.x;
			}
			Line line = new Line(k, b, denominator);
			if(map.containsKey(line)) {
				map.get(line).add(m);
				map.get(line).add(n);
			} else {
				Set<Point> set = new HashSet<>();
				set.add(m);
				set.add(n);
				map.put(line, set);
			}
		}
	}
	int max = 0;
	for(Map.Entry<Line, Set<Point>> entry : map.entrySet()) {
		max = Math.max(max, entry.getValue().size());
	}
	return max;
}

static class Line {

int k;

int b;

int denominator;

Line(int k, int b, int denominator) {

this.k = k;

this.b = b;

this.denominator = denominator;

}

@Override

public boolean equals(Object o) {

if (this == o) return true;

if (o == null || getClass() != o.getClass()) return false;

Line line = (Line) o;

if(b*line.denominator != line.b*denominator) return false;

if(k*line.denominator != line.k*denominator) return false;

return true;

}

@Override

public int hashCode() {

int t = denominator;

if(t == 0) t = 1;

int result = k/t;

result = 31 * result + b/t;

return result;

}

public static int maxPoints(Point[] points) {

if(points == null) return 0;

if(points.length == 1 || points.length == 2) return points.length;

Map<Line, Set<Point>> map = new HashMap<>();

for(int i = 0; i < points.length; i++) {

for(int j = i + 1; j < points.length; j++) {

Point m = points[i];

Point n = points[j];

int k = n.y - m.y;

int b = n.x * m.y - m.x * n.y;

int denominator = n.x - m.x;

if(denominator == 0) {

k = Integer.MAX_VALUE;

b = n.x;

}

Line line = new Line(k, b, denominator);

if(map.containsKey(line)) {

map.get(line).add(m);

map.get(line).add(n);

} else {

Set<Point> set = new HashSet<>();

set.add(m);

set.add(n);

map.put(line, set);

}

int max = 0;

for(Map.Entry<Line, Set<Point>> entry : map.entrySet()) {

max = Math.max(max, entry.getValue().size());

}

return max;

}

Trapping Rain Water

题目

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

答案

思路：若在最尾端添加一个无限高的bar，则整个题目变的十分容易：由于最后一个bar是无限高的，所以盛下水后的形状就是一个梯度上升的阶梯状，只需要扫面一遍，遇到比当前最大值小的则表示可以盛下水。现在的问题是，没有这么一块无限高的bar，那怎么办？很简单，扫面一遍，找出最高的bar，然后以此为分界线，看成是2个带有无限高bar的阶梯状就OK了。

public int trap(int[] A) {
	if(A == null || A.length < 3) return 0;

	int maxPos = 0;
	for(int i = 1; i < A.length; i++) {
		maxPos = A[i] > A[maxPos] ? i : maxPos;
	}
	int sum = 0;
	int nPos = 0;
	for(int i = 0; i < maxPos; i++) {
		if(A[i] <= A[nPos]) sum += A[nPos] - A[i];
		else nPos = i;
	}
	nPos = A.length - 1;
	for(int i = A.length - 1; i > maxPos; i--) {
		if(A[i] <= A[nPos]) sum += A[nPos] - A[i];
		else nPos = i;
	}

	return sum;
}

public int trap(int[] A) {

if(A == null || A.length < 3) return 0;

int maxPos = 0;

for(int i = 1; i < A.length; i++) {

maxPos = A[i] > A[maxPos] ? i : maxPos;

}

int sum = 0;

int nPos = 0;

for(int i = 0; i < maxPos; i++) {

if(A[i] <= A[nPos]) sum += A[nPos] - A[i];

else nPos = i;

}

nPos = A.length - 1;

for(int i = A.length - 1; i > maxPos; i--) {

if(A[i] <= A[nPos]) sum += A[nPos] - A[i];

else nPos = i;

}

return sum;

}

Word Break II

题目

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given

s = "catsanddog",

dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

答案

思路：没有思路，递归+一个Stack就可以了。时间复杂度，应该是O(n^2)？

吐槽：采用前序遍历+深度优先方式会出现Time Limit Exceeded，原因是有一个测试用例是："aaaaaaaaaaa..aaab",["a","aa","aaa","aaaa",...,"aaaaaaaaaa"]。所以采用前序深度遍历会出现超时。采用后续是不是则不会遇到这个问题？必须不是，只是刚好没有类似"baaaaaaaaaaa..aaa",["a","aa","aaa","aaaa",...,"aaaaaaaaaa"]这样的测试用例而已。

那么问题来了：要是有这样的用例该怎么办？~~（吐槽能让你吐出好程序么。。）~~

private Stack<String> pool = new Stack<>();
private List<String> list = new ArrayList<>();

public List<String> wordBreak(String s, Set<String> dict) {
	if(s == null || s.equals("")) return null;
	String word = "";

	for(int i = s.length() - 1; i > -1; i--) {
		word = s.charAt(i) + word;
		if(dict.contains(word) && i == 0) {
			String w = "";
			for(String str : pool) {
				w = str + " " + w;
			}
			w = word + " " + w;
			list.add(w.trim());
			if(!pool.isEmpty())pool.pop();
		} else if(dict.contains(word) && i != 0) {
			pool.push(word);
			wordBreak(s.substring(0, i), dict);
		} else if(!dict.contains(word) && i == 0) {
			if(!pool.isEmpty())pool.pop();
		}
	}

	return list;
}

private Stack<String> pool = new Stack<>();

private List<String> list = new ArrayList<>();

public List<String> wordBreak(String s, Set<String> dict) {

if(s == null || s.equals("")) return null;

String word = "";

for(int i = s.length() - 1; i > -1; i--) {

word = s.charAt(i) + word;

if(dict.contains(word) && i == 0) {

String w = "";

for(String str : pool) {

w = str + " " + w;

}

w = word + " " + w;

list.add(w.trim());

if(!pool.isEmpty())pool.pop();

} else if(dict.contains(word) && i != 0) {

pool.push(word);

wordBreak(s.substring(0, i), dict);

} else if(!dict.contains(word) && i == 0) {

if(!pool.isEmpty())pool.pop();

}

return list;

}

（转载本站文章请注明作者和出处程序员的自我修养 – SelfUp.cn ，请勿用于任何商业用途）

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7条评论

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2014年12月8日下午6:09

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2014年12月8日下午6:09

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2014年12月17日下午12:36

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