leetcode-python 第七周

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LeetCode Online Judge
https://leetcode.com/

1.Contain With Most Water [96ms]

#方法1：头尾指针，比较头尾两块板的高度，当low小于high的时候#如果 high--并不会增加容量，所以low++，遍历取最大值class Solution(object):    def maxArea(self, height):        low, high = 0, len(height) - 1        ans = 0        if high+1 < 2 :            return 0        while low < high :            area = (high - low) * min(height[low], height[high])            ans = max(ans, area)            if height[low] > height[high] :                high -= 1            else :                low += 1        return ans

2.Find Minimum In Rotated Array II [64ms]

#方法1：二分，不能判断就 low += 1，这里用了很多的min，应该可以优化#方法2：Java版本的，while循环限制了 nums[low] > nums[high]，默认一定是旋转的，这样可以直接返回nums[low]class Solution(object):    def findMin(self, nums):        ans = 1000000        low, high = 0, len(nums) - 1        while low < high :            mid = int((low + high) / 2)            ans = min(ans, nums[mid])            if nums[low] < nums[mid] :                ans = min(ans, nums[low])                low = mid + 1            elif nums[low] > nums[mid] :                ans = min(ans, nums[mid])                high = mid            else :                low += 1        ans = min(ans, nums[low])        return ansclass Solution {public:    int findMin(vector<int>& nums) {        if(nums.empty())            return 0;        if(nums.size() == 1)            return nums[0];        int n = nums.size();        int low = 0;        int high = n-1;        while(low < high && nums[low] >= nums[high])        {            int mid = low + (high-low)/2;            if(nums[mid] < nums[low])   // mid is in second part                high = mid;            else if(nums[mid] == nums[low])                low ++;            else                low = mid+1;        }        return nums[low];    }};

3.Find The Duplicate Number [56ms]

#方法1：回环检测法，膜拜大神，但感觉限制条件蛮多的 [56ms]#方法2：二分法加鸽巢原理 [98ms]class Solution(object):    def findDuplicate(self, nums):        # slow指针每次走一步，fast指针每次走两步        slow, fast = 0, 0        #检测回环入口，但是不知道那个是目标数        while True :            slow = nums[slow]            fast = nums[nums[fast]]            if slow == fast :                break        #找打重复的目标        find = 0        while True :            slow = nums[slow]            find = nums[find]            if slow == find :                return slowclass Solution(object):    def findDuplicate(self, nums):        lth = len(nums)        low, high = 0, lth - 1        while low <= high :            mid = int((low + high) / 2)            cnt = 0            #计算小于中间数的个数            for i in range(lth) :                if nums[i] <= mid :                    cnt += 1            #如果计算值大于中位数，则肯定在前半部分            if cnt > mid :                high = mid - 1            else :                low = mid + 1        return low

4.Insert Delete GetRandom O(1) [576ms]

#方法1：用python内置的字典结构#或者用一个数组和一个哈希表的结构#数组用来储存数和生成随机数#哈希表记录值的位置，删除需要把它换到后面#判断是否在键值内，数下键值的数目判断就可以class RandomizedSet(object):    def __init__(self):        """        Initialize your data structure here.        """        self.d = {}    def insert(self, val):        """        Inserts a value to the set. Returns true if the set did not already contain the specified element.        :type val: int        :rtype: bool        """        if not self.d.has_key(str(val)) :            self.d[str(val)] = val            return True        else :            return False    def remove(self, val):        """        Removes a value from the set. Returns true if the set contained the specified element.        :type val: int        :rtype: bool        """        if self.d.has_key(str(val)) :            del self.d[str(val)]            return True        else :            return False    def getRandom(self):        """        Get a random element from the set.        :rtype: int        """        import random        keys = list(self.d.keys())        lth = len(keys)        if lth == 0 :            return None        else :            r = random.randint(0, lth-1)            return self.d[keys[r]]# Your RandomizedSet object will be instantiated and called as such:# obj = RandomizedSet()# param_1 = obj.insert(val)# param_2 = obj.remove(val)# param_3 = obj.getRandom()

5.Longest Consecutive Sequence [40ms]

#方法1：先取出一个元素，然后向两边扩展，记得遍历完删除，保证O(n)复杂度# 用while uni 比 用for 更省事# 用set 比 用自己建的{}更省事# set有pop方法和remove方法class Solution(object):    def longestConsecutive(self, nums):        uni = set(nums)        ans = 0        while uni :            n = uni.pop()            left = n - 1            right = n + 1            count = 1            while left in uni :                uni.remove(left)                left -= 1                count += 1            while right in uni :                uni.remove(right)                right += 1                count += 1            ans = max(ans, count)        return ans

6.Next Permutation [72ms]

#方法1：参照网上的方法#注意的地方是切片的resvece方法无效和逆序的写法class Solution(object):    def nextPermutation(self, nums):        lth = len(nums)        flag = -1        #从右到左找到第一个打破升序的位置flag        for i in range(lth-1, 0, -1) :            if nums[i-1] < nums[i] :                flag = i - 1                break        if flag < 0 :            nums.reverse()            return         #从右到左找到第一个比打破位置大的数        for i in range(lth-1, -1, -1) :            if nums[i] > nums[flag] :                nums[i], nums[flag] = nums[flag], nums[i]                break        #从flag+1开始后面逆序        start, end = flag+1, lth-1        while start < end :            nums[start], nums[end] = nums[end], nums[start]            start += 1            end -= 1

7.Trapping Rain Water [57ms]

#方法1：能装多少水取决于左右最大的高度，最后判断是否为正其实是看它是不是边界 [80ms]#方法2：从两边开始搜索，次高那边往里缩，因为水面由次高决定，有最高在肯定水面能达到次高 [57ms]class Solution(object):    def trap(self, height):        lth = len(height)        maxheight = 0        leftmax = [0 for i in range(lth)]        for i in range(lth) :            leftmax[i] = maxheight            maxheight = maxheight if maxheight > height[i] else height[i]        maxheight = 0        rightmax = [0 for i in range(lth)]        for i in range(lth-1, -1, -1) :            rightmax[i] = maxheight            maxheight = maxheight if maxheight > height[i] else height[i]        water = 0        for i in range(lth) :            tmp = min(leftmax[i], rightmax[i]) - height[i]            if tmp > 0 :                water += tmp        return waterclass Solution(object):    def trap(self, height):        secHeight = 0        left, right = 0, len(height) - 1        water = 0        while left < right :            if height[left] < height[right] :                secHeight = max(height[left], secHeight)                water += secHeight - height[left]                left += 1            else :                secHeight = max(height[right], secHeight)                water += secHeight - height[right]                right -= 1        return water

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