CareerCup the Lexicographically Smallest Permutation DI序列字典序最小 LeetCode 942. DI String Match

最新推荐文章于 2024-09-23 03:49:01 发布

原创最新推荐文章于 2024-09-23 03:49:01 发布 · 915 阅读

0 ·

CC 4.0 BY-SA版权

算法同时被 2 个专栏收录

475 篇文章

订阅专栏

c++

265 篇文章

订阅专栏

本文探讨了一种算法，用于根据给定的排列签名计算出最可能的原始排列顺序。通过实例演示，展示了如何利用签名信息推断出原始数组元素的相对位置，特别关注于在遇到‘I’或‘D’时的处理策略。

You are given an array of n elements [1,2,....n]. For example {3,2,1,6,7,4,5}.
Now we create a signature of this array by comparing every consecutive pir of elements. If they increase, write I else write D. For example for the above array, the signature would be "DDIIDI". The signature thus has a length of N-1. Now the question is given a signature, compute the lexicographically smallest permutation of [1,2,....n]. Write the below function in language of your choice.

vector* FindPermute(const string& signature);

-----------------------------------------------------------------------------------------------

When meeting some 'I', we could get back to previousLow(the position of previous 'I'), the elements between previous 'I' and current 'I' are decreasing...

找到I就可以填数了，然后回填之前的D

#include<stdio.h>
#include<string>
#include<vector>
using namespace std;

class Solution {
public:
	vector<int> solve(const string& input) {
		int len = input.length(), i, j, element = 1, previousLow = -1;
		vector<int> res(len + 1, 0);
		for (i = 0; i <= len; ++i) {
			if (input[i] == 'I' || i == len) {
				for (j = i; j > previousLow; --j)
					res[j] = element++;
				previousLow = i;
			}
		}
		return res;
	}
};

int main()
{
	Solution s = Solution();
	vector<int> res1 = s.solve("DDIIDI"); //3214657
	vector<int> res2 = s.solve("DDDDDI"); //6543217
	vector<int> res3 = s.solve("DDDDDD"); //7654321
	vector<int> res4 = s.solve("IIIIII"); //1234567
	vector<int> res5 = s.solve("IIIDDD"); //1237654
	vector<int> res6 = s.solve("DDDIII"); //4321567
		
	return 0;
}

Python Version:

class Solution:
    def diStringMatch(self, S):
        l = len(S)
        res = [0 for i in range(l + 1)]
        prevI, cur = -1, 0
        for i in range(l + 1):
            if (i == l or S[i] == 'I'):
                res[i] = cur
                cur += 1
                for j in range(i-1, prevI, -1):
                    res[j] = cur
                    cur += 1
                prevI = i
        return res


s = Solution()
print(s.diStringMatch("DDI"))

Could submit it in Leetcode https://leetcode.com/problems/di-string-match/submissions/