Archive for November 18th, 2013
[leetcode] Triangle
Posted November 18, 2013
on:这题其实和robot一样,每个点取决于从上到自己和从左上角到自己哪个小,但是一变成一维dp就做了好久。主要是因为这个不是取决于头顶和左边了,而是头顶和左上角,于是需要一个临时数组。
public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) { int res = Integer.MAX_VALUE; int[] d = new int[triangle.get(triangle.size() - 1).size()]; d[0] = triangle.get(0).get(0); for (int i = 1; i < triangle.size(); i++) { int rowSize = triangle.get(i).size(); int[] tmp = new int[rowSize]; for (int j = 0; j < rowSize; j++) tmp[j] = d[j]; for (int j = 0; j < triangle.get(i).size(); j++) { int fromAbove = j < rowSize - 1 ? d[j] : Integer.MAX_VALUE; int fromUpperLeft = j > 0 ? tmp[j - 1] : Integer.MAX_VALUE; d[j] = Math.min(fromAbove, fromUpperLeft) + triangle.get(i).get(j); } } for (int i = 0; i < d.length; i++) { res = Math.min(res, d[i]); } return res; }
- In: leetcode
- 2 Comments
还是两个map,一个want, 一个has,i指当前window的开头,j指结尾;want满足了的时候就试试能不能从i处删点什么。写了25分钟然后各种debugger,这题还得重做。
public String minWindow(String S, String T) { Map<Character, Integer> has = new HashMap<Character, Integer>(); Map<Character, Integer> want = new HashMap<Character, Integer>(); String res = S + S; char[] s = S.toCharArray(); char[] t = T.toCharArray(); for (int i = 0; i < t.length; i++) { want.put(t[i], want.get(t[i]) == null ? 1 : want.get(t[i]) + 1); has.put(t[i], 0); } int i = 0; for (int j = i; j < s.length; j++) { if (want.containsKey(s[j])) { has.put(s[j], has.get(s[j]) + 1); if (satisfy(has, want)) {// satisfies everything nows, try move i while (!want.containsKey(s[i]) || has.get(s[i]) > want.get(s[i])) { //move i until not movable if (want.containsKey(s[i]) && has.get(s[i]) > want.get(s[i])) has.put(s[i], has.get(s[i]) - 1); i++; } String currWindow = S.substring(i, j + 1); res = res.length() > currWindow.length() ? currWindow : res; } } } return res.length() > S.length() ? "" : res; } private boolean satisfy( Map<Character, Integer> has, Map<Character, Integer> want) { for (Character c : want.keySet()) { if (has.get(c) < want.get(c)) return false; } return true; }
如果两个field没关系,可以生成两个锁,这样update c1必须是synchronized, c2也是。但是要是直接synchronize两个函数,说明update c1的时候c2也得干瞅着,浪费了。就跟reader writer类似,read的时候另一个reader明明也可以read的,但要是完全上锁,则read, write都要等别人了。
public class Foo {
ReentrantLock L1 = new ReentrantLock();
ReentrantLock L2 = new ReentrantLock();
int c1, c2;
public void incrementC1 {
synchronized(L1) {
c1++;
}
}
public void incrementC2 {
L2.lock();
c2++;
L2.unlock();
}
}
一种用lock,一种用synchronized object的区别就是,用lock的可以手动处理c2++ throw exception的情况,而用synchronized出了scope索就自动丢了。