A simple greedy construction procedure:
- Fill out the seq of ‘I‘ with 0, 1, 2, 3...
- Similarly, for the seq of ‘D‘, fill it out with n, n - 1, n - 2...
- At last, don‘t forget to put last element
class Solution(object): def diStringMatch(self, S): """ :type S: str :rtype: List[int] """ a, b = 0, len(S) r = [] for c in S: if c == ‘I‘: r += [a] a += 1 else: r += [b] b -= 1 return r + [a]
原文：https://www.cnblogs.com/tonix/p/10136521.html