Python 给定一个字符串，移动到所需字符的索引

关于

问题是：

给定一个长度为26的字符串键盘，指示 键盘（索引从0到25），最初您的手指位于索引0处。 要键入字符，必须将手指移到字符的食指上 理想的性格。将手指从食指i移动到食指i所需的时间 指数j为| i-j |

这是一个相当简单的问题，不应该花费太多时间。因此，我尝试了以下解决方案

def key_front(num, k, word, keyboard):
    for x in range(k):
        if word[num] == keyboard[x]:
            return x
    return 0

def calculateTime(keyboard, word):
    k = len(keyboard)
    count = 0
    i = 0
    w = len(word)
    diff = 0
    while i < w:
        front = 0
        front = key_front(i, k, word, keyboard)
        print(front,diff,word[i],count)
        if word[i] == word[i-1]:
            diff = 0
        diff = abs(front-diff)
        count += diff
        i = i + 1
    return count

这似乎有效：

keyboard = "pyevcountbgjklxfqwimrdhazs"
word = "gxafwhexxykisc...jebxjfrfoxatxhi"

keys = {char: index for index, char in enumerate(keyboard)}

index = 0
s = 0
for char in word:
    i = keys[char]
    s += abs(index - i)
    index = i
print(s)  # 66042

我首先为键盘创建查找表

键

，然后我在你的

单词

的

字符

上循环，更新

索引

我当前所在的位置，并在运行的sum

s

中计算新的ald和旧的

索引

之间的绝对差值

---

在您的解决方案中，

diff

始终指定给

0

，并且永远不会是任何其他值：您只处理当前字母和以前字母相同的情况

---

另外，您似乎没有保留当前的索引。

我对您的代码做了一点更改，但只是为了得到正确的结果。如果你想优化你的代码，@hiro progator的答案是一个很好的例子

# this function is working fine
def key_front(num, k, word, keyboard):
    for x in range(k):
        if word[num] == keyboard[x]:
            return x
    return 0

def calculateTime(keyboard, word):
    k = len(keyboard)
    count = 0
    i = 0
    w = len(word)
    diff = 0
    # I've added a new variable here to store "your finger's" previous position:
    prev_front = 0
    while i < w:
        # there is no need to initialize "front" here
        front = key_front(i, k, word, keyboard)
        print(front,diff,word[i],count)
        #if word[i] == word[i-1], front and prev_front would be the same, resulting diff == 0

        diff = front - prev_front
        # assigning our new variable prev_front
        prev_front = front
        # diff might be negative so we only need it's absolute value
        count += abs(diff)
        i = i + 1
    return count

#此函数工作正常
def键前（数字、k、字、键盘）：
对于范围（k）内的x：
如果字[num]==键盘[x]：
返回x
返回0
def计算时间（键盘、word）：
k=len（键盘）
计数=0
i=0
w=len（字）
差异=0
#我在这里添加了一个新变量来存储“你的手指”以前的位置：
前面=0
而我

那么“长度为26的字符串键盘”呢？你说你在使用递归，但我在这里没有看到任何递归函数。您的程序中的代码是否比您向我们展示的代码多？请在hashmap[word[i]时提供一个链接以查看此循环

。如果你说“我的函数是递归的，因为它有一个while循环”，那就不是递归。如果函数调用自身，则它是递归的
calculateTime
从不调用
calculateTime
，因此它不是递归的。@HiroProgator是的，它们是递归的。这不是原始问题的一部分，但更像是后续问题。假设字符是按顺序重复的，我知道了。但是你能告诉我我做错了什么吗？现在不是这样的
front
是我们正在寻找的字符，
diff
是前一个字符的索引。是的，我有。使用Dict很好。但我仍在尝试如何修改我的解决方案。谢谢你的帮助
keyboard = "pyevcountbgjklxfqwimrdhazs"
word = "gxafwhexxykisc...jebxjfrfoxatxhi"

keys = {char: index for index, char in enumerate(keyboard)}

index = 0
s = 0
for char in word:
    i = keys[char]
    s += abs(index - i)
    index = i
print(s)  # 66042

# this function is working fine
def key_front(num, k, word, keyboard):
    for x in range(k):
        if word[num] == keyboard[x]:
            return x
    return 0

def calculateTime(keyboard, word):
    k = len(keyboard)
    count = 0
    i = 0
    w = len(word)
    diff = 0
    # I've added a new variable here to store "your finger's" previous position:
    prev_front = 0
    while i < w:
        # there is no need to initialize "front" here
        front = key_front(i, k, word, keyboard)
        print(front,diff,word[i],count)
        #if word[i] == word[i-1], front and prev_front would be the same, resulting diff == 0

        diff = front - prev_front
        # assigning our new variable prev_front
        prev_front = front
        # diff might be negative so we only need it's absolute value
        count += abs(diff)
        i = i + 1
    return count