LeetCode—66、88、118、119、121 Array(Easy)

66. Plus One

Given a non-negative integer represented as a non-empty array of digits, plus one to the integer.

You may assume the integer do not contain any leading zero, except the number 0 itself.

The digits are stored such that the most significant digit is at the head of the list.

题目地址：https://leetcode.com/problems/plus-one/description/

题意：给定一个非负整数表示为非空数组，对该整数加1

思路：进位

class Solution(object):
    def plusOne(self, digits):
        """
        :type digits: List[int]
        :rtype: List[int]
        """
        carry, n = 1, len(digits)
        ans = [0 for i in range(n)]
        for i in range(n):
            ans[i] = (carry + digits[n-i-1]) % 10
            carry = (carry + digits[n-i-1]) / 10
        if carry:
            ans.append(carry)
        return ans[::-1]

 

88. Merge Sorted Array

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.

题目地址：https://leetcode.com/problems/merge-sorted-array/description/

题意：给定两个已经排序好的数值，将nums2中的前n个插入nums1中，保持顺序

思路：

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: void Do not return anything, modify nums1 in-place instead.
        """
        while m > 0 and n > 0:
            if nums1[m-1] >= nums2[n-1]:
                nums1[m+n-1] = nums1[m-1]
                m -= 1
            else:
                nums1[m+n-1] = nums2[n-1]
                n -= 1
        if n > 0:
            nums1[:n] = nums2[:n]

 

118. Pascal's Triangle

Given numRows, generate the first numRows of Pascal's triangle.

For example, given numRows = 5,
Return

[
     [1],
    [1,1],
   [1,2,1],
  [1,3,3,1],
 [1,4,6,4,1]
]

题目地址：https://leetcode.com/problems/pascals-triangle/description/

题意：给定行，返回杨辉三角

思路：

class Solution(object):
    def generate(self, numRows):
        """
        :type numRows: int
        :rtype: List[List[int]]
        """
        ans = [[1],[1,1]]
        for i in range(1, numRows):
            temp = [1] + [ans[-1][t]+ans[-1][t+1] for t in range(len(ans[-1])-1)] +[1]
            ans.append(temp)
        return ans[:numRows]

119. Pascal's Triangle II

Given an index k, return the k^th row of the Pascal's triangle.

For example, given k = 3,
Return [1,3,3,1].

Note:
Could you optimize your algorithm to use only O(k) extra space?

题目地址：https://leetcode.com/problems/pascals-triangle-ii/description/

题意：返回杨辉三角的指定行

思路：和上题差不多

class Solution(object):
    def getRow(self, rowIndex):
        """
        :type rowIndex: int
        :rtype: List[int]
        """
        ans = [[1],[1,1]]
        if rowIndex < 2:
            return ans[rowIndex]
        for i in range(1, rowIndex):
            temp = [1] + [ans[-1][t]+ans[-1][t+1] for t in range(len(ans[-1])-1)] + [1]
            ans.append(temp)
        return temp

121. Best Time to Buy and Sell Stock

Say you have an array for which the i^th element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

题目地址：https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/

题意：给定每一天的股价，求最大的利润

思路：先买入，在卖出,每次更新当前最小价格

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices:
            return 0
        ans, money, n = 0, prices[0], len(prices)
        for i in range(n):
            ans=max(ans,prices[i]-money)
            money = min(prices[i],money)
        return ans