2020-08-19

110. 平衡二叉树

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        def dfs(root):
            if root is None: return 0,True
            lh , lok = dfs(root.left)
            rh , rok = dfs(root.right)
            ok = True
            if not lok or not rok : ok = False
            if abs(rh-lh)>1: ok = False
            h = max(lh, rh)+1
            return h, ok
        
        _, ret = dfs(root)
        return ret

 

109. 有序链表转换二叉搜索树

快慢指针+二分

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def sortedListToBST(self, head):
        """
        :type head: ListNode
        :rtype: TreeNode
        """
        def build(head):
            if head is None: return None
            pre = None
            slow = head
            fast = head
            while fast and fast.next:
                pre = slow
                slow = slow.next
                fast = fast.next.next

            root = TreeNode(slow.val)

            if pre:
                pre.next = None
                root.left = build(head)
            root.right = build(slow.next)
            return root

        return build(head)

647. 回文子串

dp: dp[i][j]代表 i~j是否为回文串。

class Solution(object):
    def countSubstrings(self, s):
        """
        :type s: str
        :rtype: int
        """
        dp = []*len(s)
        for i in range(len(s)):
            dp.append([0]*len(s))
        
        ans = 0
        for l in range(1,len(s)+1):
            for i in range(len(s)):
                if i+l-1 >= len(s): continue
                if l==1: dp[i][i] = 1
                elif l==2: dp[i][i+1] = (s[i]==s[i+1])
                else : dp[i][i+l-1] = dp[i+1][i+l-2] and s[i]==s[i+l-1]
                #print(i, i+l-1)
                if dp[i][i+l-1]:
                    ans+=1

        return ans

51. N皇后

标记主副对角线 和列 进行爆搜。

class Solution(object):
    def solveNQueens(self, n):
        """
        :type n: int
        :rtype: List[List[str]]
        """
        vis_col = [0]*n
        vis_l = [0]*2*n
        vis_r = [0]*2*n
        col = []
        res = []
        
        def dfs(row):
            if row == n:
                mould = []
                for i in range(n):
                    mould.append("")
                    for j in range(n):
                        if col[i]==j:
                            mould[i] += 'Q'
                        else: 
                            mould[i] += '.'
                res.append(mould)
                return
            for i in range(n):
                if vis_col[i]==0 and vis_l[i-row+n]==0 and vis_r[i+row]==0:
                    vis_col[i] = 1
                    vis_l[i-row+n] = 1
                    vis_r[i+row] = 1
                    col.append(i)
                    dfs(row+1)
                    col.pop()
                    vis_col[i] = 0
                    vis_l[i-row+n] = 0
                    vis_r[i+row] = 0

        dfs(0)
        return res

 

61. 旋转链表

构造环形链表

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if head is None: return head
        tail = head
        p = head
        l = 0
        while p:
            tail = p
            p = p.next
            l+=1
        
        tail.next = head
        k = k%l
        k = l-k-1 # 找到前一个点
        while k:
            head = head.next
            k-=1
        newhead = head.next

        head.next = None
        return newhead