96. Unique Binary Search Trees&&95. Unique Binary Search Trees II&&241. Different Ways to Ad...

96. Unique Binary Search Trees

Given  n, how many structurally unique BST's (binary search trees) that store values 1... n?

For example,Given n = 3, there are a total of 5 unique BST's.

1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3   Tree Dynamic Programming   public class Solution { public int numTrees(int n) { //numTrees(n) = Sum(numTrees(n-1-i) * numTrees(n-1-i)), where i is [0,n-1] if(n==0) return 1; int[] nums = new int[n+1]; nums[0] = 1; nums[1] = 1; for(int i = 2; i<n+1; ++i) { int totalChildren = i - 1; for(int left = 0; left<=totalChildren;++left) { nums[i]+=nums[left]*nums[totalChildren-left]; } } return nums[n]; /* public int numTrees(int n) { if(n==0) return 1; if(n <= 2) return n; int total = 0; for(int i = 1; i<=n;++i) { int left = numTrees(i-1); int right = numTrees(n-i); total += left * right; } return total; } */ } }

 

95. Unique Binary Search Trees II

Given an integer n, generate all structurally unique BST's (binary search trees) that store values 1...n.

Given n = 3, your program should return all 5 unique BST's shown below.

1 3 3 2 1 \ / / / \ \ 3 2 1 1 3 2 / / \ \ 2 1 2 3   Tree Dynamic Programming   /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<TreeNode> generateTrees(int n) { return generateTrees(1,n); } private List<TreeNode> generateTrees(int from, int to) { List<TreeNode> results = new ArrayList<TreeNode>(); if(from > to) return results; if(from == to){ results.add(new TreeNode(from)); return results; } for(int i = from; i<=to; ++i){ List<TreeNode> left = generateTrees(from, i-1); List<TreeNode> right = generateTrees(i+1, to); if(left.size() == 0) { for(TreeNode r:right){ TreeNode root = new TreeNode(i); root.right = r; results.add(root); } } else if(right.size() == 0) { for(TreeNode l:left){ TreeNode root = new TreeNode(i); root.left = l; results.add(root); } } else{ for(TreeNode l:left){ for(TreeNode r:right){ TreeNode root = new TreeNode(i); root.left = l; root.right = r; results.add(root); } } } } return results; } }

 

241. Different Ways to Add Parentheses

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.

Example 1

Input: "2-1-1".

((2-1)-1) = 0 (2-(1-1)) = 2

Output: [0, 2]

Example 2

Input: "2*3-4*5"

(2*(3-(4*5))) = -34 ((2*3)-(4*5)) = -14 ((2*(3-4))*5) = -10 (2*((3-4)*5)) = -10 (((2*3)-4)*5) = 10

Output: [-34, -14, -10, -10, 10]

Divide and Conquer class Solution { public List<Integer> diffWaysToCompute(String input) { List<Integer> ret = new LinkedList<>(); for (int i = 0; i < input.length(); ++i) { char c = input.charAt(i); if (c == '-' || c == '*' || c == '+') { for (int left : diffWaysToCompute(input.substring(0, i))) { for (int right : diffWaysToCompute(input.substring(i + 1))) { if (c == '+') ret.add(left + right); else if (c == '-') ret.add(left - right); else if (c == '*') ret.add(left * right); } } } } if (ret.size() == 0) ret.add(Integer.valueOf(input)); return ret; } }

 

282. Expression Add Operators

Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or * between the digits so they evaluate to the target value.

Examples: 

"123", 6 -> ["1+2+3", "1*2*3"] "232", 8 -> ["2*3+2", "2+3*2"] "105", 5 -> ["1*0+5","10-5"] "00", 0 -> ["0+0", "0-0", "0*0"] "3456237490", 9191 -> []   Divide and Conquer   public class Solution { public List<String> addOperators(String num, int target) { List<String> ret = new ArrayList<>(); if (num == null || num.length() == 0) return ret; helper(ret, "", 0, num, target, 0, 0); return ret; } /** * Iterate through all possibilities to add operators to str, * and add to results if target is achieved. */ private void helper(List<String> ret, String currentString, long currentResult, String str, int target, int startInd, long previousNum) { if (startInd == str.length()) { if (target == currentResult) ret.add(currentString); return; } for (int endInd = startInd; endInd < str.length(); ++endInd) { //0 cannot be a starting number, unless it is 0 by itself if (str.charAt(startInd) == '0' && endInd > startInd) //e.g. "105", "05" might be treated as 5 //so it will errorly return ["1*0+5","1*5","10-5"] break; Long num = Long.parseLong(str.substring(startInd, endInd + 1)); if (startInd == 0) helper(ret, num.toString(), num, str, target, endInd + 1, num); else { helper(ret, currentString + "+" + num, currentResult + num, str, target, endInd + 1, num); helper(ret, currentString + "-" + num, currentResult - num, str, target, endInd + 1, -num); //if previous number is used as a multiplier, we need to remove it from earlier sum long product = previousNum * num; helper(ret, currentString + "*" + num, currentResult - previousNum + product, str, target, endInd + 1, product); } } } }

 

 

   

转载于:https://www.cnblogs.com/neweracoding/p/5722323.html