Archive for November 18th, 2013
[leetcode] Triangle
Posted on: November 18, 2013
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;
}
Tags: 2WayDP
- 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。
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