标签：span ret 写法 sort err etc sum void har
public interface Intervals { /** * Adds an interval [from, to] into internal structure. */ void addInterval(int from, int to); /** * Returns a total length covered by intervals. * If several intervals intersect, intersection should be counted only once. * Example: * * addInterval(3, 6) * addInterval(8, 9) * addInterval(1, 5) * * getTotalCoveredLength() -> 6 * i.e. [1,5] and [3,6] intersect and give a total covered interval [1,6] * [1,6] and [8,9] don‘t intersect so total covered length is a sum for both intervals, that is 6. * * _________ * ___ * ____________ * * 0 1 2 3 4 5 6 7 8 9 10 * */ int getTotalCoveredLength(); }
版上常见的那个interval题目， 要实现一个接口，实现addInterval和getTotalCoverage两个函数。 基本算是leetcode的两个interval题目的变种， 第一遍bug free写了一个add O(n)和get O(1)的方法， 然后交流了一下之后做了些小优化。 followup是问有没有让add方法更efficient的写法, 然后我说了下add O(1), get O(nlogn)的方法， 然后又讨论了一些优化的问题
// add Time O(1), get Time O(nlgn)
class MyIntervals {
List<Length> l = new LinkedList<Length>();
public void addInterval(int from, int to) {
l.add(new Length(from,to));
}
public int getTotalCoveredLength(){
if (l == null || l.size() == 0) {
return 0;
}
Comparator<Length> comp = new Comparator<Length>() {
@Override
public int compare(Length o1, Length o2) {
return o1.x == o2.x ? o1.y - o2.y : o1.x - o2.x;
}
};// ";"不能漏
Collections.sort(l, comp);
int ans = 0;
ans += l.get(0).y - l.get(0).x;
for (int i = 1; i < l.size(); i++) {
if (l.get(i).y <= l.get(i - 1).y) {
continue;
}
if (l.get(i).x < l.get(i - 1).y) {
ans += l.get(i).y - l.get(i - 1).y;
} else {
ans += l.get(i).y - l.get(i).x;
}
}
return ans;
}
}
class Length{
public int x, y;
public Length(int x, int y) {
this.x = x;
this.y = y;
}
}
// add Time O(n), get Time O(1)
class MyIntervals {
List<int[]> intervals = new LinkedList<int[]>();
int sum = 0;
public void addInterval(int from, int to) {
if (intervals == null) {
int[] temp = new int[]{from, to};
sum = to - from;
intervals.add(temp);
return;
}
List<int[]> result = new LinkedList<int[]>();
int insertPos = 0;
int len = 0;
int[] newInterval = new int[]{from, to};
for (int[] cur : intervals) {
if (cur[1] < newInterval[0]) {
result.add(cur);
len += cur[1] - cur[0];
insertPos++;
} else if (cur[0] > newInterval[1]) {
result.add(cur);
len += cur[1] - cur[0];
} else {
newInterval[0] = Math.min(newInterval[0], cur[0]);
newInterval[1] = Math.max(cur[1], newInterval[1]);
}
}
len += newInterval[1] - newInterval[0];
result.add(insertPos, newInterval);
intervals = result;
sum = len;
return;
}
public int getTotalCoveredLength() {
return sum;
}
}
标签：span ret 写法 sort err etc sum void har
原文地址：http://www.cnblogs.com/apanda009/p/7927904.html
