LeetCode 763. Partition Labels

https://leetcode.com/problems/partition-labels/description/

A string S of lowercase letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts. 

Example 1:

Input: S = "ababcbacadefegdehijhklij"
Output: [9,7,8]
Explanation:
The partition is "ababcbaca", "defegde", "hijhklij".
This is a partition so that each letter appears in at most one part.
A partition like "ababcbacadefegde", "hijhklij" is incorrect, because it splits S into less parts.

Note:

1. S will have length in range [1, 500].
2. S will consist of lowercase letters ('a' to 'z') only.

---

• 字符串中等题。滑动窗口+贪心。时间复杂度O(N)，空间复杂度O(N)。
• 首先依然是做hash，不过这次是对所有char记录它们最后出现的位置。
• 然后定义三个指针，window开始left和结束right，当前i。每次移动i，然后判断当前S[i]最后出现的位置是否大于当前right，如果是，则扩展right为last[S[i]]。
• 如果当前i == window结束right，则代表当前window中所有的字符都只在当前window中出现。所以可以把当前window的size记录。同时移动left到下一个window的起点。
• https://leetcode.com/problems/partition-labels/solution/

//
//  main.cpp
//  LeetCode
//
//  Created by Hao on 2017/3/16.
//  Copyright © 2017年 Hao. All rights reserved.
//

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

class Solution {
public:
    vector<int> partitionLabels(string S) {
        vector<int> vResult;
        
        // corner case
        if (S.empty()) return vResult;
        
        unordered_map<char, int> last;
        
        // the last occurred index of S[i]
        for (int i = 0; i < S.size(); ++ i)
            last[S.at(i)] = i;
        
        // start and end of the current window
        int left = 0, right = 0;
        
        // current pointer
        for (int i = 0; i < S.size(); ++ i) {
            right = max(right, last[S.at(i)]); // maximum index of the current window
            
            // reach the end of current window
            if (i == right) {
                vResult.push_back(right - left + 1); // size of the current window
                left = i + 1; // move to start of the next window
            }
        }
        
        return vResult;
    }
};

int main(int argc, char* argv[])
{
    Solution    testSolution;
    
    vector<string>  inputs = {"", "ababcbacadefegdehijhklij", "abcdefg", "aaaaaaaaaa"};
    vector<int>     result;
    
    /*
     
     9 7 8
     1 1 1 1 1 1 1
     10
     */
    for (auto input : inputs) {
        result = testSolution.partitionLabels(input);
        
        for (auto it : result)
            cout << it << " ";
        cout << endl;
    }

    return 0;
}