149-153

149： 直线最多的点数 

思路就是 对于每一个点 看后面能和他组成的直线数
相同的直线怎么判断呢 就用一个字典 标签是 '1' +str((Y2-Y1)/(X2-X1)) 就是用横坐标归一化
不过要注意 平行于X轴和Y轴的直线要单独标出来 我用的是X0 和 Y0
判断完一个数后 字典要重置

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        n = len(points)
        linedict = {'X0':0,'Y0':0}
        max = 0
        for i in range(n):
            for j in range(i+1,n):
                if points[j][1] - points[i][1] == 0:
                    linedict['X0'] += 1
                if points[j][0] - points[i][0] == 0:
                    linedict['Y0'] += 1
                elif points[j][1] - points[i][1] != 0:
                    strtemp = '1' + str((points[j][1] - points[i][1])/(points[j][0] - points[i][0]))
                    if strtemp in linedict:
                        linedict[strtemp] += 1
                    else:
                        linedict[strtemp] = 1
            for each in linedict:
                if linedict[each] > max:
                    max = linedict[each]
            linedict = {'X0':0,'Y0':0}
        return max +1

作者：yizhu-jia
链接：https://leetcode-cn.com/problems/max-points-on-a-line/solution/bu-cuo-bu-cuo-yi-ci-ti-jiao-jiu-cheng-go-cm81/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

 

150 ： 逆波兰式

出题人把答案都写出来了  看不起谁呢？？？？

这道题麻烦的是正负整数的取整

class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        nums = []
        ope = []
        for each in tokens:
            if len(each) >=2 or ord(each) >=48 and ord(each) <=57  :
                nums.append(int(each))
            else:
                temp1 = nums.pop()
                temp2 = nums.pop()
                if each == '+':
                    nums.append(temp2+temp1)
                elif each == '-':
                    nums.append(temp2-temp1)
                elif each == '*':
                    nums.append(temp2*temp1)
                elif each == '/':
                    if temp2*temp1 < 0:
                        nums.append(-1*(abs(temp2)//abs(temp1)))
                    else:
                        nums.append(temp2//temp1)
        return nums[0]


作者：yizhu-jia
链接：https://leetcode-cn.com/problems/evaluate-reverse-polish-notation/solution/kan-bu-qi-shui-ni-chu-ti-huan-ba-da-an-x-lhjp/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

 

151：

反转字符串里的单词  

想了半天 被9成人击败 

class Solution:
    def reverseWords(self, s: str) -> str:
        s = s[::-1]
        n = len(s)
        def findnextletter(index):
            n = len(s)
            while index < n and s[index] == ' ':
                index += 1
            letterstart = index              #找到下一個單詞開始
            while index < n and s[index] != ' ':
                index += 1
            if len(s) > n :
                index += 1
            return letterstart,index        #找到下一個單詞結束
        start = 0

        while start < n :
            letterstart,letterend = findnextletter(start)
            s = s[:start]+s[letterstart:letterend][::-1]+' '+s[letterend:]
            start = letterend - (letterstart-start) + 1


        return s.rstrip(' ')

作者：yizhu-jia
链接：https://leetcode-cn.com/problems/reverse-words-in-a-string/solution/zen-yao-hui-shi-xiang-liao-ban-tian-bei-69i21/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

 

152：

乘积最大子数组 

能用多少方法用多少種 我就是要过 好像提交了无数次 

主要思路就是 乘到绝对值最大时 如果负数的个数是偶数个就不管  如果是奇数个 就除以绝对值最小的那个 。

 

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        if nums == [0]:
            return 0
        negcount = 0
        n = len(nums)
        maxmul = float('-inf')
        curmul = 1
        gate = False
        firneg = 1
        lastneg = 0
        numcount = 0
        for i,each in enumerate(nums):
            if each == 0 :
                if gate:
                    if negcount % 2 == 1 and numcount != 1:
                        curmul = curmul/max(firneg,lastneg)        #这里是除以绝对值更小的那个
                    if curmul > maxmul:
                        maxmul = curmul
                    if 0 > maxmul:
                        maxmul = 0
                    firneg = 1
                    lastneg = 0
                    curmul = 1
                    negcount = 0
                    numcount = 0
            else:
                numcount += 1
                gate = True
                if negcount == 0:
                    firneg *= each
                curmul *= each
                lastneg *= each
                if each < 0:
                    negcount += 1
                    lastneg = each
                if i == n-1:
                    if negcount % 2 == 1 and numcount != 1:
                        curmul = curmul/max(firneg,lastneg)        #这里是除以绝对值更小的那个
                    if curmul > maxmul:
                        maxmul = curmul

        return int(maxmul)

153：

寻找旋转排序中最小值 

我用的是二分 结果被大部分击败  我怀疑他们min 

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left = 0
        n = len(nums)
        right = n-1
        while left <= right:
            mid = (left +right) //2
            if mid == n-1 or mid == 0:
                return min(nums[left],nums[right])
            if nums[mid] < nums[mid+1] and nums[mid] < nums[mid-1]:
                return nums[mid]
            if nums[mid] > nums[left]:
                if nums[mid] > nums[right]:
                    left = mid + 1
                else:
                    return nums[left]
            else:        #当小于左端点时。
                if nums[mid] > nums[right]:
                    left = mid + 1
                else:
                    right = mid -1


作者：yizhu-jia
链接：https://leetcode-cn.com/problems/find-minimum-in-rotated-sorted-array/solution/wei-sha-zi-wo-zhe-yao-man-er-qie-kong-ji-ltks/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。