Leetcode 算法题07

169. Majority Element

输入一个列表，找出其中出现次数超过列表长度一半的数

我的代码：

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        count = collections.Counter(nums)
        for i in count:
            if count[i] > len(nums)/2:
                return i

大神的代码：既然占了代码一半，排序后中间的那个数肯定就是所求

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return sorted(nums)[len(nums)/2]

167. Two Sum II - Input array is sorted

给一个排好序的列表和目标数，找出列表中相加得到目标数的两个数的索引

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

我的代码：第一次超时了，没有记录，这次是想了个不超时的思路，写得多了点

class Solution(object):
    def twoSum(self, numbers, target):
        """
        :type numbers: List[int]
        :type target: int
        :rtype: List[int]
        """
        number = collections.Counter(numbers)
        print(number)
        keys = list(number.keys())
        keys.sort()
        print(keys)
        for i in range(len(keys)):
            if keys[i] > target:
                return False
            add = 0
            while i+add <len(keys) and keys[i+add] + keys[i] < target:
                add += 1
            if i+add < len(keys) and keys[i+add] + keys[i] == target:
                if add != 0:
                    ans1 = sum(list(map(lambda x:number[x],keys[:i])))+1
                    ans2 = sum(list(map(lambda x:number[x],keys[:i+add])))+1
                    return [ans1,ans2]
                elif add == 0 and number[keys[i]]>1:
                    ans1 = sum(list(map(lambda x:number[x],keys[:i])))+1
                    ans2 = ans1+1
                    return [ans1,ans2]

大神的代码：自愧不如

# two-pointer
def twoSum1(self, numbers, target):
    l, r = 0, len(numbers)-1
    while l < r:
        s = numbers[l] + numbers[r]
        if s == target:
            return [l+1, r+1]
        elif s < target:
            l += 1
        else:
            r -= 1
 
# dictionary           
def twoSum2(self, numbers, target):
    dic = {}
    for i, num in enumerate(numbers):
        if target-num in dic:
            return [dic[target-num]+1, i+1]
        dic[num] = i
 
# binary search        
def twoSum(self, numbers, target):
    for i in xrange(len(numbers)):
        l, r = i+1, len(numbers)-1
        tmp = target - numbers[i]
        while l <= r:
            mid = l + (r-l)//2
            if numbers[mid] == tmp:
                return [i+1, mid+1]
            elif numbers[mid] < tmp:
                l = mid+1
            else:
                r = mid-1

387. First Unique Character in a String

输入一个字符串，找到第一个在字符串中没有重复出现过的字母的索引

Examples:

s = "leetcode"
return 0.

s = "loveleetcode",
return 2.

我的代码：

class Solution(object):
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        a=collections.Counter(s)
        ans = []
        for i in a:
            if a[i] == 1:
                ans.append(i)
        if ans == []:
            return -1
        return min([s.find(j) for j in ans])

其他思路：

def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        letters='abcdefghijklmnopqrstuvwxyz'
        index=[s.index(l) for l in letters if s.count(l) == 1]
        return min(index) if len(index) > 0 else -1

237. Delete Node in a Linked List

删除节点

我的代码：感觉有时候出现的问题好蠢

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        node.val ,node.next= node.next.val,node.next.next