Python任务支付代码出现list index out of range错误求助

问题描述 #

编写用于计算任务分段支付的Python代码时，遭遇list index out of range（列表索引越界）错误，以下是相关代码、输入及报错信息，需要排查并解决该问题。

原代码 #

total_task=float(input("Enter the assigned total task length(in half-hour(s)):"))
total_len=total_task*2
leng=int(total_len)
payments=[]
hours=[]
for i in range(leng):
    print("Enter the payment value( in TL) for task portion ID ",(i+1)," having length ",((i+1)*0.5)," hour(s):")
    portionLen=int(input())
    payments.append(portionLen)
    hours.append(portionLen)
paymentsTable=[]
for i in range(leng):
    paymentsRow=[]
    for j in range(leng):
        paymentsRow.append(0)
    paymentsTable.append(paymentsRow)
for i in range(leng):
    paymentsTable[i][i]=payments[i]
for i in range(leng):
    for j in range(1,leng+1):
        maxPayment=0
        for k in range(j):
            pay=paymentsTable[i][k]+paymentsTable[k+1][j]
            if(pay>maxPayment):
                maxPayment=pay
        paymentsTable[i][j]=maxPayment
idTable=[]
for i in range(leng):
    idTableRow=[]
    for j in range(leng):
        idTableRow.append(0)
    idTable.append(idTableRow)
for i in range(leng):
    idTable[i][i]=i+1
for i in range(leng):
    for j in range(1,leng+1):
        maxPayment=0
        for k in range(j):
            pay = paymentsTable[i][k] + paymentsTable[k + 1][j]
            if (pay > maxPayment):
                maxPayment = pay
        paymentsTable[i][j] = maxPayment
for i in range(leng):
    for j in range(1,leng+1):
        maxPayment=0
        for k in range(j):
            pay = paymentsTable[i][k] + paymentsTable[k + 1][j]
            if (pay > maxPayment):
                maxPayment = pay
                idTable[i][j]=k+1

示例输入 #

Enter the assigned total task length(in half-hour(s)):**2**
Enter the payment value( in TL) for task portion ID  1  having length  0.5  hour(s):
**100**
Enter the payment value( in TL) for task portion ID  2  having length  1.0  hour(s):
**400**
Enter the payment value( in TL) for task portion ID  3  having length  1.5  hour(s):
**500**
Enter the payment value( in TL) for task portion ID  4  having length  2.0  hour(s):
**600**

报错信息 #

line 23, in <module>
    pay=paymentsTable[i][k]+paymentsTable[k+1][j]
IndexError: list index out of range

问题分析 #

1. 索引越界核心原因：
   原代码中paymentsTable和idTable被创建为leng x leng的二维列表（索引范围0到leng-1），但后续循环中j的取值范围是range(1, leng+1)，当j=leng时，访问paymentsTable[k+1][j]会超出列表的最大索引（leng-1），直接触发索引越界错误。
2. 输入逻辑错误：
   hours.append(portionLen)这行代码错误地将支付金额添加到了时长列表hours中，实际应该添加对应分段的时长(i+1)*0.5。
3. 冗余代码：
   原代码重复了三次几乎完全相同的paymentsTable计算循环，属于无意义的冗余，既浪费资源又增加维护成本。

解决方案 #

1. 调整二维列表维度：
   将paymentsTable和idTable创建为(leng+1) x (leng+1)的二维列表，这样索引可以覆盖到j=leng的情况，匹配后续循环的取值范围。
2. 修正输入逻辑：
   修改hours列表的赋值语句，添加正确的分段时长。
3. 移除冗余代码：
   合并重复的计算逻辑，只保留一次核心的动态规划计算过程。

修改后的完整代码 #

total_task = float(input("Enter the assigned total task length(in half-hour(s)):"))
total_len = total_task * 2
leng = int(total_len)
payments = []
hours = []

# 修正输入逻辑，正确记录分段时长和支付金额
for i in range(leng):
    portion_hour = (i + 1) * 0.5
    print(f"Enter the payment value( in TL) for task portion ID {i+1} having length {portion_hour} hour(s):")
    portion_pay = int(input())
    payments.append(portion_pay)
    hours.append(portion_hour)

# 创建(leng+1)x(leng+1)的二维列表，避免索引越界
paymentsTable = [[0]*(leng+1) for _ in range(leng+1)]
# 初始化对角线值
for i in range(1, leng+1):
    paymentsTable[i][i] = payments[i-1]

# 动态规划计算最大支付，只保留一次核心循环
for length in range(2, leng+1):  # length表示子任务的分段数
    for i in range(1, leng - length + 2):
        j = i + length - 1
        maxPayment = 0
        for k in range(i, j):
            current_pay = paymentsTable[i][k] + paymentsTable[k+1][j]
            if current_pay > maxPayment:
                maxPayment = current_pay
        paymentsTable[i][j] = maxPayment

# 创建(leng+1)x(leng+1)的idTable
idTable = [[0]*(leng+1) for _ in range(leng+1)]
# 初始化对角线值
for i in range(1, leng+1):
    idTable[i][i] = i

# 重新计算并记录分割点
for length in range(2, leng+1):
    for i in range(1, leng - length + 2):
        j = i + length - 1
        maxPayment = 0
        for k in range(i, j):
            current_pay = paymentsTable[i][k] + paymentsTable[k+1][j]
            if current_pay > maxPayment:
                maxPayment = current_pay
                idTable[i][j] = k

# 可以添加结果输出，例如打印最大支付和分割点
print(f"Maximum total payment: {paymentsTable[1][leng]} TL")
print("Task split points:", idTable[1][leng])

内容的提问来源于stack exchange，提问作者DonKirsolft