同向双指针模板——里面if j == n 和 sum >= s的细节要区分哪

406. 和大于S的最小子数组

中文

English

给定一个由 n 个正整数组成的数组和一个正整数 s ，请找出该数组中满足其和 ≥ s 的最小长度子数组。如果无解，则返回 -1。

样例

样例 1:

输入: [2,3,1,2,4,3], s = 7
输出: 2
解释: 子数组 [4,3] 是该条件下的最小长度子数组。

样例 2:

输入: [1, 2, 3, 4, 5], s = 100
输出: -1

挑战

如果你已经完成了O(nlogn)时间复杂度的编程，请再试试 O(n)时间复杂度。

class Solution:
    """
    @param nums: an array of integers
    @param s: An integer
    @return: an integer representing the minimum size of subarray
    """
    def minimumSize2(self, nums, s):
        # write your code here
        ans = float('inf')
        n = len(nums)

        j = 0
        s2 = 0
        for i in range(n):
            while j < n and s2 + nums[j] < s:
                s2 += nums[j]
                j += 1

            if j == n:
                break

            ans = min(ans, j - i + 1)

            # s2 += nums[j]
            s2 -= nums[i]


        return ans if ans != float('inf') else -1
        
    
    def minimumSize(self, nums, s):
        # write your code here
        ans = float('inf')
        n = len(nums)

        j = 0
        s2 = 0
        for i in range(n):
            while j < n and s2 < s:
                s2 += nums[j]
                j += 1

            if s2 >= s:
                ans = min(ans, j - i)

            s2 -= nums[i]

        return ans if ans != float('inf') else -1

 

同向双指针， 模版1是强化班侯卫东老师介绍的 模版2是高频班老顽童老师介绍的

# 模版1
class Solution:
    """
    @param nums: an array of integers
    @param s: An integer
    @return: an integer representing the minimum size of subarray
    """
    def minimumSize(self, nums, s):
        # write your code here
        
        
        left, right = 0, 0 
        
        n = len(nums)
        
        target = s 
        
        addup = 0 
        
        ans = sys.maxsize 
        
        for left in range(n):
            
            while right < n and addup < target:
                
                addup += nums[right]
                
                right += 1 
                
            if addup >= target: # 满足条件
                
                ans = min(right - left, ans)
                
            addup -= nums[left]
            
            
        return -1 if ans == sys.maxsize else ans 

# 模版2: 枚举右端点，左端点不回头
class Solution:
    """
    @param nums: an array of integers
    @param s: An integer
    @return: an integer representing the minimum size of subarray
    """                
            
        
    def minimumSize(self, nums, s):
        
        ans = sys.maxsize 
        
        left = 0 
        
        addup = 0 
        
        for right in range(len(nums)):
            
            addup += nums[right]
            
            while addup >= s:
                
                ans = min(ans, right - left + 1)
                
                addup -= nums[left]
                
                left += 1 
                
        return ans if ans != sys.maxsize else -1