标签：算法 string break 思想 null div highlight eof parent
241. Different Ways to Add Parentheses (Medium)
Input: "2-1-1".
((2-1)-1) = 0
(2-(1-1)) = 2
Output : [0, 2]
public List<Integer> diffWaysToCompute(String input) {
List<Integer> ways = new ArrayList<>();
for (int i = 0; i < input.length(); i++) {
char c = input.charAt(i);
if (c == ‘+‘ || c == ‘-‘ || c == ‘*‘) {
List<Integer> left = diffWaysToCompute(input.substring(0, i));
List<Integer> right = diffWaysToCompute(input.substring(i + 1));
for (int l : left) {
for (int r : right) {
switch (c) {
case ‘+‘:
ways.add(l + r);
break;
case ‘-‘:
ways.add(l - r);
break;
case ‘*‘:
ways.add(l * r);
break;
}
}
}
}
}
if (ways.size() == 0) {
ways.add(Integer.valueOf(input));
}
return ways;
}
95. Unique Binary Search Trees II (Medium)
给定一个数字 n，要求生成所有值为 1...n 的二叉搜索树。
Input: 3
Output:
[
[1,null,3,2],
[3,2,null,1],
[3,1,null,null,2],
[2,1,3],
[1,null,2,null,3]
]
Explanation:
The above output corresponds to the 5 unique BST‘s shown below:
1 3 3 2 1
\ / / / \ 3 2 1 1 3 2
/ / \ 2 1 2 3
public List<TreeNode> generateTrees(int n) {
if (n < 1) {
return new LinkedList<TreeNode>();
}
return generateSubtrees(1, n);
}
private List<TreeNode> generateSubtrees(int s, int e) {
List<TreeNode> res = new LinkedList<TreeNode>();
if (s > e) {
res.add(null);
return res;
}
for (int i = s; i <= e; ++i) {
List<TreeNode> leftSubtrees = generateSubtrees(s, i - 1);
List<TreeNode> rightSubtrees = generateSubtrees(i + 1, e);
for (TreeNode left : leftSubtrees) {
for (TreeNode right : rightSubtrees) {
TreeNode root = new TreeNode(i);
root.left = left;
root.right = right;
res.add(root);
}
}
}
return res;
}
标签：算法 string break 思想 null div highlight eof parent
原文地址：https://www.cnblogs.com/daimasanjiaomao/p/11009084.html
