leetcode-python 第十周

LeetCode Online Judge
https://leetcode.com/

1. Three Sum [252ms]

# 方法1：先确定一个变量i，然后用两个指针搜索# 注意两次去重class Solution(object):    def threeSum(self, nums):        """        :type nums: List[int]        :rtype: List[List[int]]        """        if len(nums) < 3: return []        ans = []        nums.sort()        i = 0        while i < len(nums) - 2:            print(i)            start, end = i + 1, len(nums) - 1            while start < end:                if nums[i] + nums[start] + nums[end] == 0:                    ans.append([nums[i], nums[start], nums[end]])                    # 去重                    while start < end and nums[start] == nums[start+1]:                        start += 1                    start += 1                elif nums[i] + nums[start] + nums[end] < 0:                    start += 1                else:                    end -= 1            # 去重            while i+1 < len(nums) and nums[i] == nums[i+1]:                i += 1            i += 1        return ans 

2.Longest Substring Without Repeating Characters

# 方法1：双重循环，一定超时# 方法2：头指针当遇到重复时就往后缩，顺便删除记录# 尾指针当遇到不同元素时往后，还有就是更新完毕也要+1class Solution(object):    def lengthOfLongestSubstring(self, s):        if len(s) < 2:            return len(s)        ans = 0        d = {}        low = 0        d[s[low]] = 0        tmp = 1        high = low + 1        while high < len(s):            if s[high] in d.keys():                ans = max(ans, tmp)                index = d[s[high]]                while low <= index:                    del d[s[low]]                    tmp -= 1                    low += 1                d[s[high]] = high                high += 1                tmp += 1            else:                d[s[high]] = high                tmp += 1                high += 1        return max(ans, tmp)

3. Maximum Product Subarray [64ms]

# 方法1：除了保存最大数，还要保存一个最小数，防止忽略了负数乘负数class Solution(object):    def maxProduct(self, nums):        """        :type nums: List[int]        :rtype: int        """        if len(nums) < 2 :            return nums[0]        dpmax = [nums[i] for i in range(len(nums))]        dpmin = [nums[i] for i in range(len(nums))]        ans = nums[0]        for i in range(1, len(nums)) :            dpmin[i] = min(dpmin[i], dpmin[i-1] * nums[i], dpmax[i-1] * nums[i])            dpmax[i] = max(dpmax[i], dpmax[i-1] * nums[i], dpmin[i-1] * nums[i])            ans = max(ans, dpmax[i])        return ans

4.Median of Two Sorted Arrays [148ms]

# http://blog.csdn.net/yutianzuijin/article/details/11499917/class Solution(object):    def findMedianSortedArrays(self, nums1, nums2):        """        :type nums1: List[int]        :type nums2: List[int]        :rtype: float        """        lth = len(nums1) + len(nums2)        mid = int(lth / 2)        if lth & 0x1:            return self.findKth(nums1, nums2, mid + 1) # 奇数，终止条件为k==1        else:            return (self.findKth(nums1, nums2, mid) + self.findKth(nums1, nums2, mid + 1)) / 2. # 偶数    def findKth(self, nums1, nums2, k):        if len(nums1) > len(nums2):            return self.findKth(nums2, nums1, k) # 默认nums1的长度小于nums2        if len(nums1) == 0:            return nums2[k-1]         if k == 1:            return min(nums1[0], nums2[0])        sub1 = int(min(k / 2, len(nums1)))        sub2 = k - sub1        if nums1[sub1 - 1] < nums2[sub2 - 1]:            return self.findKth(nums1[sub1:], nums2, k - sub1)        elif nums1[sub1 - 1] > nums2[sub2 - 1]:            return self.findKth(nums1, nums2[sub2:], k - sub2)        else:            return nums1[sub1 - 1]

5. Sprial Matrix [64ms]

class Solution(object):    def spiralOrder(self, matrix):        """        :type matrix: List[List[int]]        :rtype: List[int]        """        ans = []        if len(matrix) < 1 :            return ans        if len(matrix) == 1 and len(matrix[0]) == 1 :            ans.append(matrix[0][0])            return ans        cnt = 0        clen = len(matrix) - 1        rlen = len(matrix[0]) - 1        lth = len(matrix) * len(matrix[0])        c = 0        r = 0        while True :            for j in range(r, r+rlen) :                ans.append(matrix[c][j])                cnt += 1                if cnt >= lth : return ans            for i in range(c, c+clen) :                ans.append(matrix[i][r+rlen])                cnt += 1                if cnt >= lth : return ans            for j in range(r+rlen, r, -1) :                ans.append(matrix[c+clen][j])                cnt += 1                if cnt >= lth : return ans            for i in range(c+clen, c, -1) :                ans.append(matrix[i][r])                cnt += 1                if cnt >= lth : return ans            c += 1            r += 1            clen -= 2            rlen -= 2            if rlen < 0 :                rlen = 0            if clen < 0 :                clen = 0            if c + clen == c and r == r + rlen:                ans.append(matrix[c+clen][c])                cnt += 1                if cnt >= lth : return ans        return ans

6. Two Sum II [64ms]

# 方法1：双指针法class Solution(object):    def twoSum(self, numbers, target):        """        :type numbers: List[int]        :type target: int        :rtype: List[int]        """        low, high = 0, len(numbers) - 1        while low < high:            if numbers[low] + numbers[high] < target:                low += 1            elif numbers[low] + numbers[high] > target:                high -= 1            else:                return [low + 1, high + 1]

7.Valid Anagram [104ms]

# 方法1：哈希表class Solution(object):    def isAnagram(self, s, t):        """        :type s: str        :type t: str        :rtype: bool        """        if len(s) != len(t):            return False        cnt = {}        for s_i in range(len(s)):            try:                cnt[s[s_i]] += 1            except:                cnt[s[s_i]] = 1        for t_i in range(len(t)):            try:                cnt[t[t_i]] -= 1                if cnt[t[t_i]] == 0:                    del cnt[t[t_i]]            except:                return False        return True