96. Unique Binary Search Trees 等题
2016-11-25 14:30
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96. Unique Binary Search Trees
原题: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
题解:
给定一个正整数n,问数字1~n可以构成多少棵不同的二叉搜索树。
首先要知道二叉搜索树的特性,就是每个节点的左子树中的所有元素都比它小,右子树中的节点都比它大。而且两个子树都是二叉搜索树。那么可以这样说。在1~n数字中,如果数字
i作为根节点,那么根节点的左子树一定是由
1 ~ i-1构成,右子树一定是
i+1 ~ n。在这种情况下,
i+1 ~ n中的数字可以整体减去i,变成
1 ~ n-i的情况。很明显就是DP的状态转移了。
因此用
f(n)表示n个数字有多少种不同的二叉搜索树,有:
# python表示 f(0) = f(1) = 1 f(n) = sum([f(i-1)*f(n-i) for i in range(1, n+1)])
代码:
class Solution { public: int numTrees(int n) { if(n<2) { return 1; } vector<int> f(n+1); f[0] = 1; f[1] = 1; for(int i=1; i<=n; i++) { f[i] = 0; for(int j=1; j<=i; j++) { f[i] += f[j-1] * f[i-j]; } } return f ; } };
416. Partition Equal Subset Sum
原题:Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.
Note:
Each of the array element will not exceed 100.
The array size will not exceed 200.
Example 1:
Input: [1, 5, 11, 5] Output: true Explanation: The array can be partitioned as [1, 5, 5] and [11].
Example 2:
Input: [1, 2, 3, 5] Output: false Explanation: The array cannot be partitioned into equal sum subsets.
题解:
给一组正整数,问能否把它们分成两个子集,使得两个集合的和相等。
首先可以排除一个明显无解的情况:数组的和为奇数。
把数组和除于二后,得到一个
subsum,于是把问题变成了
subset sum问题。
subset sum问题有伪多项式的解法,用
f(i, j)表示0到i个元素能否找到子集的和为j。
状态转移式为:
f(i, j) = (i == j) || (f(i-1, j)) || (j-i>=0 && f(i-1, j-i))
代码:
class Solution { public: bool canPartition(vector<int>& nums) { int sum = 0; for(auto x: nums) { sum += x; } if(sum % 2) { return false; } sum /= 2; vector<vector<bool>> f(nums.size(), vector<bool>(sum+1)); // f[0][sum] = (nums[0] == sum); for(int j=1; j<=sum; ++j) { f[0][j] = (nums[0] == j); } for(int i=1; i<nums.size(); ++i) { for(int j=1; j<=sum; ++j) { f[i][j] = (nums[i] == j) || (f[i-1][j]) || (j-nums[i]>=0 && f[i-1][j-nums[i]]); } } return f[nums.size()-1][sum]; } };
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