lintcode题目记录4

Russian Doll Envelopes    Largest Divisible Subset     Two Sum - Input array is sorted

Russian Doll Envelopes

俄罗斯玩偶嵌套问题，这个是典型的dp问题···强行遍历会提示超时，然而整了好久也没整明白怎么整，网上搜了下 把问题归结为求最长递增子序列问题··然而本人愚钝还是想不明白为啥可以这样做··虽然出来的结果是对的·····

先把数据排序， 用python内建的排序函数进行排序，但是因为当x相等时，y要按从大到小拍，所以要传一个cmp进去，python3.x不支持cmp了 所以 用了一个转换，转换成key，如果直接key设置为x默认y会按从小到大拍

这样算的结果是对的·但是那个迭代的dp不是一个有效的序列···但是长度是对的···

class Solution:
    # @param {int[][]} envelopes a number of envelopes with widths and heights
    # @return {int} the maximum number of envelopes
    def maxEnvelopes(self, envelopes):
        # Write your code here
        import functools
        nums = sorted(envelopes,key= functools.cmp_to_key(lambda x,y:x[0]-y[0] if x[0] != y[0] else y[1] - x[1]))
        size = len(nums)
        dp = []
        for x in range(size):
            low, high = 0, len(dp) - 1
            while low <= high:
                mid = (low + high)//2
                if dp[mid][1] < nums[x][1]:
                    low = mid + 1
                else:
                    high = mid - 1
            if low < len(dp):
                dp[low] = nums[x]
            else:
                dp.append(nums[x])
        return len(dp)

Largest Divisible Subset

标签写的是动态规划 ·感觉没啥规划还是直接强行遍历的··

class Solution:
    # @param {int[]} nums a set of distinct positive integers
    # @return {int[]} the largest subset
    def largestDivisibleSubset(self, nums):
        # Write your code here
        n=len(nums)
        nums= sorted(nums,reverse=True)
        res=[]
        res.append([nums[0]])
        for i in range(1,n):
            cur=nums[i]
            for r in res:
                if  r[-1] % cur== 0:
                    r.append(cur)
            if i==1:res.append([nums[0]])
            res.append([nums[i]])
        res=sorted(res,key=lambda x:len(x),reverse=True)
        return res[0]

Two Sum - Input array is sorted

有序数组，找出一个组合之和是给定目标值，题目写的有序基本就是说用二分查找吧···，而且还要求index2>index1相当简单了··，遍历一遍就可以了··

class Solution:
    """
    @param nums {int[]} n array of Integer
    @param target {int} = nums[index1] + nums[index2]
    @return {int[]} [index1 + 1, index2 + 1] (index1 < index2)
    """
    def twoSum(self, nums, target):
        # Write your code here
        index1=0
        index2=-1
        for i in range(len(nums)-1):
            index1 = i
            index2 = -1
            start = i+1
            end = len(nums)-1
            st = target - nums[index1]
            while start <= end:
                mid = (start + end) // 2
                if nums[mid] < st:
                    start = mid + 1
                elif nums[mid] > st:
                    end = mid - 1
                else:
                    index2 = mid
                    return [index1 + 1, index2 + 1]