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LeetCode 每日一题 2022/8/29-2022/9/4

news 来源：原创 2024/5/12 3:57:42

记录了初步解题思路以及本地实现代码；并不一定为最优也希望大家能一起探讨一起进步

8/29 1470. 重新排列数组

重新排序 x0,xn,x1,xn+1 …

def shuffle(nums, n):
    """
    :type nums: List[int]
    :type n: int
    :rtype: List[int]
    """
    ans = [0]*(2*n)
    for i in range(n):
        ans[i*2] = nums[i]
        ans[i*2+1] = nums[i+n]
    return ans

8/30 998. 最大二叉树 II

当前节点小于val 则val为根节点当前为左子节点
当前节点大于val 则val递归入当前节点右子树中

class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
        
def insertIntoMaxTree(root, val):
    """
    :type root: TreeNode
    :type val: int
    :rtype: TreeNode
    """
    pre,cur = None,root
    while cur:
        if val > cur.val:
            node = TreeNode(val,cur,None)
            if not pre:
                return node
            pre.right = node
            return root
        else:
            pre = cur
            cur = cur.right
    pre.right = TreeNode(val)
    return root

8/31 946. 验证栈序列

将当前push入栈
判断栈顶元素是否与pop相同
如果相同则考虑下一个
如果不同则继续push

def validateStackSequences(pushed, popped):
    """
    :type pushed: List[int]
    :type popped: List[int]
    :rtype: bool
    """
    l = []
    n = len(pushed)
    loc = 0
    for p in popped:
        if l and l[-1]==p:
            l.pop()
            continue
        tag = False
        while loc<n:
            if pushed[loc]==p:
                loc+=1
                tag = True
                break
            l.append(pushed[loc])
            loc+=1
        if not tag:
            return False
    return True

9/1 1475. 商品折扣后的最终价格

从后往前考虑
stack 保存非递减序列
当前价格可以在stack中找到第一个小于等于它的数

def finalPrices(prices):
    """
    :type prices: List[int]
    :rtype: List[int]
    """
    stack = []
    n = len(prices)
    for i,v in enumerate(prices[::-1]):
        diff = 0
        while stack:
            if stack[-1]>v:
                stack.pop()
            else:
                diff = stack[-1]
                break
        stack.append(v)
        prices[n-1-i] = v-diff
    return prices

9/2 687. 最长同值路径

check用来判断当前节点能够返回给父节点最多的节点数
分别考虑左右子树
l,r为左右子树能够提供给当前节点的节点数
如果子节点与当前节点相同则提供的节点数可用curl=l curr=r
更新路径为左右两边节点和 curl+curr
能够提供给父节点的只能是一侧加上自己 max(curl,curr)+1

class TreeNode(object):
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def longestUnivaluePath(root):
    """
    :type root: TreeNode
    :rtype: int
    """
    global ans
    ans = 0
    def check(node):
        if not node:
            return 0
        global ans
        l,r = 0,0
        curl,curr = 0,0
        if node.left:
            l = check(node.left)
            if node.val==node.left.val:
                curl = l
        if node.right:
            r = check(node.right)
            if node.val==node.right.val:
                curr = r
        ans = max(ans,curl+curr)
        return max(curl,curr)+1
    check(root)
    return ans

9/3 646. 最长数对链

记录每一个出现的第一个数字后能够有的最小数
check检验第二个数为num后最多能接几个数对

def findLongestChain(pairs):
    """
    :type pairs: List[List[int]]
    :rtype: int
    """
    m = {}
    l = []
    
    for a,b in pairs:
        if a not in m:
            l.append(a)
            m[a] = b
        elif m[a]>b:
            m[a] = b
    l.sort()
    mem = {}
    def check(num):
        ans = 1
        if num in mem:
            return mem[num]
        for s in l:
            if s>num:
                ans = max(ans,check(m[s])+1)
        mem[num] = ans
        return ans
    ans = 0
    for s in l:
        ans = max(ans,check(m[s]))
    return ans