LeetCode 每日一题 2023/7/17-2023/7/23

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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7/17 415. 字符串相加

从尾至前依次相加

def addStrings(num1, num2):
    """
    :type num1: str
    :type num2: str
    :rtype: str
    """
    ans = ""
    add = 0
    n,m=len(num1),len(num2)
    i,j=n-1,m-1
    while i>=0 or j>=0 or add!=0:
        x,y = 0,0
        if i>=0:
            x = int(num1[i])
        if j>=0:
            y = int(num2[j])
        
        res = x+y+add
        add = res//10
        ans = str(res%10)+ans
        i,j=i-1,j-1
    return ans

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7/18 1851. 包含每个查询的最小区间

将区间起始位置和查询数都从小到大排序 不影响结果
区间[l,r]使用(v,r)表示 v为区间长度 r为右侧坐标
使用小顶堆来存储当前已加入区间
对于第i个区间intervals[i]左侧断点小于查询的数值x则将其加入小顶堆
如果堆顶右侧端点小于x则将其弹出

def minInterval(intervals, queries):
    """
    :type intervals: List[List[int]]
    :type queries: List[int]
    :rtype: List[int]
    """
    import heapq
    n,m=len(intervals),len(queries)
    intervals.sort()
    li = sorted((x,i) for i,x in enumerate(queries))
    ans = [-1]*m
    h = []
    i =0
    for x,ind in li:
        while i<n and intervals[i][0]<=x:
            a,b = intervals[i]
            heapq.heappush(h,(b-a+1,b))
            i+=1
        while h and h[0][1]<x:
            heapq.heappop(h)
        if h:
            ans[ind]=h[0][0]
    return ans

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7/19 874. 模拟行走机器人

模拟四个方向 判断每次走一步是否遇到障碍物
如果障碍物则不动

def robotSim(commands, obstacles):
    """
    :type commands: List[int]
    :type obstacles: List[List[int]]
    :rtype: int
    """
    step = [[0,1],[1,0],[0,-1],[-1,0]]
    d = 0
    cx,cy=0,0
    m = set([(x[0],x[1]) for x in obstacles])
    ans = 0
    for c in commands:
        if c<0:
            if c==-1:
                d+=1
            else:
                d-=1
            d%=4
        else:
            for i in range(c):
                if (cx+step[d][0],cy+step[d][1]) in m:
                    break
                cx,cy=cx+step[d][0],cy+step[d][1]
                ans = max(ans,cx*cx+cy*cy)
    return ans

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7/20 918. 环形子数组的最大和

子数组有两种情况
一种是不成环 在数组内部 该子数组为最大子数组和
一种是成环 子数组在数组头部和尾部 这中间部分为数组最小子数组和
记录最大子数组和maxs，最小子数组和mins
ans = max(maxs,total-mins)

def maxSubarraySumCircular(nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    total = sum(nums)
    curmax,curmin = 0,0
    maxs,mins = nums[0],nums[0]
    for num in nums:
        curmax = max(curmax+num,num)
        curmin = min(curmin+num,num)
        
        maxs = max(maxs,curmax)
        mins = min(mins,curmin)
    if maxs>0:
        return max(maxs,total-mins)
    else:
        return maxs

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7/21 1499. 满足不等式的最大值

yi+yj+|xi-xj|=(-xi+yi)+(xj+yj)
list存储(-x+y,x)的值
从头遍历points 去除不满足xj-xi<=k的点

def findMaxValueOfEquation(points, k):
    """
    :type points: List[List[int]]
    :type k: int
    :rtype: int
    """  
    import collections
    ans = float("-inf")
    l = collections.deque()
    for x,y in points:
        while l and x-l[0][1]>k:
            l.popleft()
        if l:
            ans = max(ans,x+y+l[0][0])
        while l and y-x>=l[-1][0]:
            l.pop()
        l.append([y-x,x])
    return ans
            

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7/22 860. 柠檬水找零

依次遍历
如果收到10元判断是否有5元
如果收到20元判断是否有10+5或者5*3

def lemonadeChange(bills):
    """
    :type bills: List[int]
    :rtype: bool
    """
    five,ten = 0,0
    for v in bills:
        if v==5:
            five+=1
        elif v==10:
            if five>0:
                five-=1
                ten+=1
            else:
                return False
        else:
            if ten>0 and five>0:
                ten-=1
                five-=1
            elif five>=3:
                five-=3
            else:
                return False
    return True

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7/23 42. 接雨水

将水装满
从左到右 记录遇到的最高的墙 让水往左流走
从右往左 记录遇到的最高的墙 让水往右流走
计算最后留下的水

def trap(height):
    """
    :type height: List[int]
    :rtype: int
    """
    n=len(height)
    ans = [float("inf")]*n
    left = 0
    right = 0
    for i in range(n):
        ans[i] = max(height[i],min(ans[i],left))
        ans[n-1-i] = max(height[n-1-i],min(ans[n-1-i],right))
        left = max(left,height[i])
        right = max(right,height[n-1-i])
    return sum(ans)-sum(height)

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