leetcode 942. DI String Match ( Python )

描述

Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length.

Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1:

If S[i] == "I", then A[i] < A[i+1]
If S[i] == "D", then A[i] > A[i+1]
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Example 1:

Input: "IDID"
Output: [0,4,1,3,2]
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Example 2:

Input: "III"
Output: [0,1,2,3]
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Example 3:

Input: "DDI"
Output: [3,2,0,1]
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Note:

1 <= S.length <= 10000
S only contains characters "I" or "D".
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解析

根据题意，找出规律，若是是 "I" 那就是找一个可选值的最小值，若是是 "D" 那就是找一个可选值的最大值，直到找完全部的值。时间复杂度为 O(N)，空间复杂度为 O(N)。

解答

class Solution(object):
def diStringMatch(self, S):
    """
    :type S: str
    :rtype: List[int]
    """
    result = []
    arr = [i for i in range(len(S)+1)]
    i = 0
    j = len(arr)-1
    for c in S:
        if c=='I':
            result.append(arr[i])
            i+=1
        else:
            result.append(arr[j])
            j-=1
    result.append(arr[i])
    return result 
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运行结果

Runtime: 60 ms, faster than 46.12% of Python online submissions for DI String Match.
Memory Usage: 13.1 MB, less than 44.62% of Python online submissions for DI String Match.
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