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数组-简单

2020-11-21 23:02:21  阅读:11  来源: 互联网

标签:startswith cur List len 数组 简单 new array


1、面试题 17.10. 主要元素 https://leetcode-cn.com/problems/find-majority-element-lcci/

考点:

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        eles = set(nums)
        all_length = len(nums)
        for ele in eles:
            if nums.count(ele) >= all_length/2:
                return ele
        return -1

2、面试题4. 二维数组中的查找

class Solution:
    def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
        if not matrix:
            return False
        row = len(matrix)
        col = len(matrix[0])
        if row == 1 and col == 0:
            return False
        
        data = []
        for i in range(row):
            data.extend(matrix[i])
        data = set(data)
        return target in data

3、面试题 16.06. 最小差

class Solution:
    def smallestDifference(self, a: List[int], b: List[int]) -> int:
        if not a or not b:
            return None
        a.sort()
        b.sort()
        len_a = len(a)
        len_b = len(b)

        cur_a = 0
        cur_b = 0
        
        min_abs = 2147483647 * 2
        while cur_a < len_a and cur_b < len_b:
            if abs(a[cur_a] - b[cur_b]) < min_abs:
                min_abs = abs(a[cur_a] - b[cur_b])
            
            if a[cur_a] > b[cur_b]:
                cur_b += 1
            else:
                cur_a += 1
        return min_abs

4、面试题 16.10. 生存人数

考点:

1、想到有一道坐公交的题,和这个类似,用一个数组表示所有年份,然后一个人一生在所有年份加1; 最后就可以直接返回 人数最多年份

class Solution:
    def maxAliveYear(self, birth: List[int], death: List[int]) -> int:
        years = [0] * 101
        for i in range(len(birth)):
            b = birth[i] - 1900
            d = death[i] - 1900
            
            for k in range(b, d + 1):
                years[k] += 1
        # print(years)
        return years.index(max(years)) + 1900

5、面试题 16.20. T9键盘

考点:

1、逆向思维,由 字符窜 反推 数字组合

class Solution:
    def getValidT9Words(self, num: str, words: List[str]) -> List[str]:
        keyb = {"a":2, "b":2, "c":2, "d":3, "e":3, "f":3, "g":4, "h":4, "i":4, "j":5, "k":5, "l":5, "m":6, "n":6, "o":6, "p":7, "q":7, "r":7, "s":7, "t":8, "u":8, "v":8, "w":9, "x":9, "y":9, "z":9}

        result = []
        for word in words:
            num_list = ""
            for ch in word:
                num_list += str(keyb[ch])
            if num_list in num:
                result.append(word)
        return result

6、面试题 17.05. 字母与数字

考点:

1、前缀和,相对正常前缀和,拐了个弯,按照数字是+1,字母是-1,求数组的前缀和

2、数组最后出现位置计算,先把数组求reverse().再使用index计算

3、注意,本题有个坑是,数字可能是多个数字组合的字符串,需要按下图绿色处理

class Solution:
    def findLongestSubarray(self, array: List[str]) -> List[str]:
        array_len = len(array)
        new_array = [0] * (array_len +1)
        cur = 0
        for i in range(array_len):
            #数字+1, 字母-1
            if array[i].startswith("1") or array[i].startswith("0") or array[i].startswith("2") or array[i].startswith("3") or array[i].startswith("4") or array[i].startswith("5") or array[i].startswith("6") or array[i].startswith("7") or array[i].startswith("8") or array[i].startswith("9"):
                cur += 1
            else:
                cur += -1
            new_array[i+1] = cur
        
        new_array_reverse = list(new_array)
        new_array_reverse.reverse()

        # 值相等,表示中间结果值为0
        max_len = 0
        start = 0
        end = 0
        matched = []
        for i in range(array_len):
            try:
                # if new_array[i] in matched:
                #     continue
                r_value = array_len - new_array_reverse.index(new_array[i])
                matched.append(new_array[i])
                if r_value -i-1  > max_len:
                    max_len = r_value - i -1
                    start = i
                    end = r_value
            except Exception as e:
                continue
        return array[start: end]

  其实按照上面做,会出现超时,几个用例跑不完,加入如下优化进行剪枝

class Solution:
    def findLongestSubarray(self, array: List[str]) -> List[str]:
        array_len = len(array)
        new_array = [0] * (array_len +1)
        cur = 0
        for i in range(array_len):
            #数字+1, 字母-1
            if array[i].startswith("1") or array[i].startswith("0") or array[i].startswith("2") or array[i].startswith("3") or array[i].startswith("4") or array[i].startswith("5") or array[i].startswith("6") or array[i].startswith("7") or array[i].startswith("8") or array[i].startswith("9"):
                cur += 1
            else:
                cur += -1
            new_array[i+1] = cur
        
        new_array_reverse = list(new_array)
        new_array_reverse.reverse()

        # 值相等,表示中间结果值为0
        max_len = 0
        start = 0
        end = 0
        matched = []
        for i in range(array_len):
            try:
                if new_array[i] in matched:   # 同一个前缀和,前面是按照最前面和最后面计算,肯定比现在计算的长,所有不用重复计算
                     continue
                r_value = array_len - new_array_reverse.index(new_array[i])
                matched.append(new_array[i])
                if r_value -i-1  > max_len:
                    max_len = r_value - i -1
                    start = i
                    end = r_value
            except Exception as e:
                continue
        return array[start: end]

  

标签:startswith,cur,List,len,数组,简单,new,array
来源: https://www.cnblogs.com/pyclq/p/14017760.html

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