算法：最大子序和笔试题

Dear 丶 2022-12-29 06:42 32阅读 0赞

### 最大子序和 ###

*  题目描述
 *  代码
 *  程序执行过程

# 题目描述 #

给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。

示例:  
输入: \[-2,1,-3,4,-1,2,1,-5,4\]  
输出: 6  
解释: 连续子数组 \[4,-1,2,1\] 的和最大，为 6。

# 代码 #

import pysnooper
    
    class Solution:
        @pysnooper.snoop()#pysnooper是装饰器，可以修改函数，作用是打印函数的参数
        def maxSubArray(self , nums ):
            # write code here
            long=len(nums)
            new_array = nums[0]
            temp = nums[0]
            for i in range(1,long):
                new_array = max(nums[i],new_array + nums[i])
                temp = max(new_array, temp)
            result =temp
            return result
    
    nums = [-2,1,-3,4,-1,2,1,-5,4]
    s = Solution()
    print(s.maxSubArray(nums))

# 程序执行过程 #

Starting var:.. nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]
    23:20:47.207620 call         5     def maxSubArray(self , nums ):
    23:20:47.207620 line         7         long=len(nums)
    New var:....... long = 9
    23:20:47.207620 line         8         new_array = nums[0]
    New var:....... new_array = -2
    23:20:47.207620 line         9         temp = nums[0]
    New var:....... temp = -2
    23:20:47.207620 line        10         for i in range(1,long):
    New var:....... i = 1
    23:20:47.207620 line        11             new_array = max(nums[i],new_array + nums[i])
    Modified var:.. new_array = 1
    23:20:47.207620 line        12             temp = max(new_array, temp)
    Modified var:.. temp = 1
    23:20:47.207620 line        10         for i in range(1,long):
    Modified var:.. i = 2
    23:20:47.207620 line        11             new_array = max(nums[i],new_array + nums[i])
    Modified var:.. new_array = -2
    23:20:47.207620 line        12             temp = max(new_array, temp)
    23:20:47.208598 line        10         for i in range(1,long):
    Modified var:.. i = 3
    23:20:47.208598 line        11             new_array = max(nums[i],new_array + nums[i])
    Modified var:.. new_array = 4
    23:20:47.208598 line        12             temp = max(new_array, temp)
    Modified var:.. temp = 4
    23:20:47.208598 line        10         for i in range(1,long):
    Modified var:.. i = 4
    23:20:47.208598 line        11             new_array = max(nums[i],new_array + nums[i])
    Modified var:.. new_array = 3
    23:20:47.208598 line        12             temp = max(new_array, temp)
    23:20:47.208598 line        10         for i in range(1,long):
    Modified var:.. i = 5
    23:20:47.208598 line        11             new_array = max(nums[i],new_array + nums[i])
    Modified var:.. new_array = 5
    23:20:47.208598 line        12             temp = max(new_array, temp)
    Modified var:.. temp = 5
    23:20:47.208598 line        10         for i in range(1,long):
    Modified var:.. i = 6
    23:20:47.208598 line        11             new_array = max(nums[i],new_array + nums[i])
    Modified var:.. new_array = 6
    23:20:47.208598 line        12             temp = max(new_array, temp)
    Modified var:.. temp = 6
    23:20:47.208598 line        10         for i in range(1,long):
    Modified var:.. i = 7
    23:20:47.208598 line        11             new_array = max(nums[i],new_array + nums[i])
    Modified var:.. new_array = 1
    23:20:47.208598 line        12             temp = max(new_array, temp)
    23:20:47.208598 line        10         for i in range(1,long):
    Modified var:.. i = 8
    23:20:47.208598 line        11             new_array = max(nums[i],new_array + nums[i])
    Modified var:.. new_array = 5
    23:20:47.208598 line        12             temp = max(new_array, temp)
    23:20:47.208598 line        10         for i in range(1,long):
    23:20:47.208598 line        13         result =temp
    New var:....... result = 6
    23:20:47.208598 line        14         return result
    23:20:47.208598 return      14         return result
    Return value:.. 6
    Elapsed time: 00:00:00.000978
    6