LeetCode 每日一题 2023/8/28-2023/9/3

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

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8/28 57. 插入区间

找到新区间起点位置 和终点位置对应的区间位置

def insert(intervals, newInterval):
    """
    :type intervals: List[List[int]]
    :type newInterval: List[int]
    :rtype: List[List[int]]
    """
    ret = []
    if len(intervals)==0:
        ret.append(newInterval)
        return ret
    
    now = newInterval
    for i in range(len(intervals)):
        v = intervals[i]
        if v[1] < now[0]:
            ret.append(v)
            continue
        elif now[1] < v[0]:
            ret.append(now)
            ret.extend(intervals[i:])
            break
        now[0] = min(v[0],now[0])
        now[1] = max(v[1],now[1])
        
    if len(ret)==0 or now[0] > ret[-1][1]:
        ret.append(now)
    return ret

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8/29 823. 带因子的二叉树

将数值从小到大排序
dp[i] 代表以arr[i]为根节点能够得到的树个数
从小到大考虑

def numFactoredBinaryTrees(arr):
    """
    :type arr: List[int]
    :rtype: int
    """
    MOD = 10**9+7
    arr.sort()
    n = len(arr)
    dp = [1]*n
    idx = {arr[i]:i for i in range(n)}
    for i,v in enumerate(arr):
        for j in range(i):
            x = arr[j]
            if x*x>v:
                break
            if x*x==v:
                dp[i]= (dp[i]+dp[j]*dp[j])%MOD
                break
            if v%x==0 and v//x in idx:
                dp[i]=(dp[i]+dp[j]*dp[idx[v//x]]*2)%MOD
    return sum(dp)%MOD

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8/30 1654. 到家的最少跳跃次数

BFS 标记当前往前为1 往后为-1
如果a>b 当前位置超过x+b之后必定再也到不了x
如果a

def minimumJumps(forbidden, a, b, x):
    """
    :type forbidden: List[int]
    :type a: int
    :type b: int
    :type x: int
    :rtype: int
    """
    s = set(forbidden)
    ans = 0
    l = [(0,1)]
    mem={(0,1)}
    while l:
        tmp = []
        for loc,k in l:
            if loc==x:
                return ans
            nxt = [(loc+a,1)]
            if k==1:
                nxt.append((loc-b,-1))
            for i,k in nxt:
                if i not in s and 0<=i<6000 and (i,k) not in mem:
                    tmp.append((i,k))
                    mem.add((i,k))
        ans+=1
        l=tmp
    return -1

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8/31 1761. 一个图中连通三元组的最小度数

m[i][j]记录i,j的连通情况
deg[i] 记录i的出度

def minTrioDegree(n, edges):
    """
    :type n: int
    :type edges: List[List[int]]
    :rtype: int
    """
    deg = [0]*n
    m = [[0]*n for i in range(n)]
    for x,y in edges:
        x,y = x-1,y-1
        m[x][y] = m[y][x] = 1
        deg[x]+=1
        deg[y]+=1
        
    ans = float("inf")
    for i in range(n):
        for j in range(i+1,n):
            if m[i][j]==1:
                for k in range(j+1,n):
                    if m[i][k]==m[j][k]==1:
                        ans = min(ans,deg[i]+deg[j]+deg[k]-6)
    return -1 if ans==float("inf") else ans

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9/1 2240. 买钢笔和铅笔的方案数

遍历可以买钢笔的个数 累加各个情况下可以买铅笔的个数

def waysToBuyPensPencils(total, cost1, cost2):
    """
    :type total: int
    :type cost1: int
    :type cost2: int
    :rtype: int
    """
    ans = 0
    for i in range(total//cost1):
        ans += (total -i*cost1)//cost2+1
    return ans

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9/2 2511. 最多可以摧毁的敌人城堡数目

从头遍历记录前一个自己城堡位置x 和 空位置y

def captureForts(forts):
    """
    :type forts: List[int]
    :rtype: int
    """
    x,y=-1,-1
    ans = 0
    for i,v in enumerate(forts):
        if v==-1:
            if x>-1:
                ans = max(i-1-x,ans)
            x,y=-1,i
        elif v==1:
            if y>-1:
                ans = max(i-1-y,ans)
            x,y=i,-1
    return ans

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9/3 1921. 消灭怪物的最大数量

计算每个怪物到城市的时间 从小到大排序
遍历每个怪物是否能在其到达前被消灭

def eliminateMaximum(dist, speed):
    """
    :type dist: List[int]
    :type speed: List[int]
    :rtype: int
    """
    n = len(dist)
    t=[0]*n
    for i in range(n):
        t[i] = dist[i]*1.0/speed[i]
    t.sort()
    ans = 0
    for i,v in enumerate(t):
        if i<v:
            ans+=1
        else:
            break
    return ans

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