Knowledge Reserve - Basic Algorithms - Substring

1. Substring

1.1 Question 560 - The subarray whose sum is k

Given an integer array  nums and an integer  k , count and return  the number of consecutive subarrays  in the array whose sum is k  .

Example 1:

Input: nums = [1,1,1], k = 2
 Output: 2

At first, I wanted to use a sliding window, but during the operation, I encountered many examples that were difficult to handle and it was difficult to handle. I looked directly at the analysis and decided to use ij range traversal, but the result was a timeout.

class Solution(object):
    def subarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        n = len(nums)
        cur_arr = []
        result = 0
        cur_sum = -100000000
        last_sum = 0
        sum_nums = 0
        for i in range(n):
            sum_nums += nums[i]
        cur_sum = sum_nums
        sum_temp = sum_nums
        for i in range(n):
            if i != 0:
                sum_temp -= nums[i-1]
            cur_sum = sum_temp
            for j in reversed(range(i,n)):
                if cur_sum == k:
                    result += 1
                cur_sum -= nums[j]

        return result

After reading the analysis, it's really hard to think about it.

class Solution(object):
    def subarraySum(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        n = len(nums)
        cur_arr = []
        result = 0
        last_sum = 0
        sum_nums = 0
        # 必须初始化为0：1，不然sum_nums-k=0时，result就不能加一了
        dic = {0:1}
        for i in range(n):
            # 求前缀和
            sum_nums += nums[i]
            # 如果字典中存在sum_nums-k，则说明sum_nums对应的i与sum_nums-k对应的i之间和为k
            # 如果sum_nums-k的键值不唯一，就表示有多个i的前缀和为sum_nums-k
            # 这时result就加上键值
            result += dic.get(sum_nums-k, 0)
            # 最后把该i的前缀和加入到字典中，若已存在则键值加一
            dic[sum_nums] = dic.get(sum_nums, 0) + 1
            print(dic, result)

        return result

1.2 Question 239 - Sliding window maximum

You are given an array of integers  numswith a  k sliding window of size moving from the far left of the array to the far right of the array. k You can only see the numbers within the sliding window  . The sliding window only moves to the right one bit at a time.

Returns  the maximum value in a sliding window  .

Thoughts: The idea seems to be quite simple, just write the sliding window directly, but it times out.

class Solution(object):
    def maxSlidingWindow(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        """
        result = []
        cur_win = []
        for i in range(k-1):
            cur_win.append(nums[i])
        for i in range(len(nums)-k+1):
            cur_max = -100000
            cur_win.append(nums[i+k-1])
            for j in range(k):
                if cur_max < cur_win[j]:
                    cur_max = cur_win[j]
            # cur_max = max(cur_win)
            result.append(cur_max)
            cur_win = cur_win[1:]

        return result

The parsing is a bit complicated.