Divide and Conquer
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2 218 The Skyline Problem 最大堆 遍历节点
public List<int[]> getSkyline(int[][] buildings) {
List<int[]> res = new ArrayList<>();
List<int[]> point = new ArrayList<>();
for (int i = 0; i < buildings.length; i++)
{
point.add(new int[]{buildings[i][0], buildings[i][2]});
point.add(new int[]{buildings[i][1], -buildings[i][2]});
}
Collections.sort(point, (a,b)-> {
if (a[0] != b[0]) return a[0] - b[0];
return b[1] - a[1];});
PriorityQueue<Integer> maxheap = new PriorityQueue<>((a,b)->(b - a));
int cur = 0, pre = 0;
for (int i = 0; i < point.size(); i++)
{
int[] b = point.get(i);
if (b[1] > 0)
{
maxheap.add(b[1]);
cur = maxheap.peek();
}
else
{
maxheap.remove(-b[1]);
cur = maxheap.peek() == null ? 0 : maxheap.peek();
}
if (pre != cur)
{
res.add(new int[]{b[0], cur});
pre = cur;
}
}
return res;
}
3 241 Different Ways to Add Parentheses 找符号,分开
public List<Integer> diffWaysToCompute(String input) {
List<Integer> res = new LinkedList<>();
for (int i = 0; i < input.length(); i++)
{
char c = input.charAt(i);
if (c == '+' || c == '-' || c == '*')
{
String a = input.substring(0, i);
String b = input.substring(i+1);
List<Integer> a1 = diffWaysToCompute(a);
List<Integer> b1 = diffWaysToCompute(b);
for (int x : a1)
{
for (int y : b1)
{
if (c == '+')
{
res.add(x + y);
}
else if (c == '-')
{
res.add(x - y);
}
else if (c == '*')
{
res.add(x * y);
}
}
}
}
}
if (res.size() == 0) res.add(Integer.valueOf(input));
return res;
}
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