C++ 解法, 执行用时: 4ms, 内存消耗: 780KB, 提交时间: 2022-07-08

static const auto io_sync_off = [](){
    std::ios::sync_with_stdio(false);
    std::cout.tie(nullptr);
    std::cin.tie(nullptr);
    return nullptr;
}();
class Solution {
public:
    vector<int> splitString(string s) {
        int h[26]={0};
        for(int i=0;i<s.size();i++)
        {
            h[s[i]-'a']=i;
        }
        vector<int>res;
        int l=0,r=0;
        for(int i=0;i<s.size();i++)
        {
            r=max(r,h[s[i]-'a']);
            if(i==r)
            {
                res.push_back(r-l+1);
                l=r+1;
            }
        }
        return res;
    }
};

C++ 解法, 执行用时: 4ms, 内存消耗: 804KB, 提交时间: 2022-08-06

static const auto io_sync_off = [](){
    std::ios::sync_with_stdio(false);
    std::cout.tie(nullptr);
    std::cin.tie(nullptr);
    return nullptr;
}();
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param s string字符串 
     * @return int整型vector
     */
    vector<int> splitString(string s) {
        // write code here
        int h[26]={0};
        for(int i=0;i<s.size();i++)
        {
            h[s[i]-'a']=i;
        }
        vector<int> res;
        int l=0,r=0;
        for(int i=0;i<s.size();i++)
        {
            r=max(r,h[s[i]-'a']);
            if(i==r)
            {
                res.push_back(r-l+1);
                l=r+1;
            }
        }
        return res;
    }
};

Go 解法, 执行用时: 4ms, 内存消耗: 1164KB, 提交时间: 2022-05-29

package main
// import "fmt"

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param s string字符串 
 * @return int整型一维数组
*/
func splitString( s string ) []int {
    // write code here
    partition := make([]int, 0, 16)
    lastPos := [26]int{}
    for i, c := range s {
        lastPos[c-'a'] = i
    }
    start, end := 0, 0
    for i, c := range s {
        if lastPos[c-'a'] > end {
            end = lastPos[c-'a']
        }
        if i == end {
            partition = append(partition, end-start+1)
            start = end + 1
        }
    }
    return partition
}

func partitionLabels(s string) (partition []int) {
    lastPos := [26]int{}
    for i, c := range s {
        lastPos[c-'a'] = i
    }
    start, end := 0, 0
    for i, c := range s {
        if lastPos[c-'a'] > end {
            end = lastPos[c-'a']
        }
        if i == end {
            partition = append(partition, end-start+1)
            start = end + 1
        }
    }
    return
}

C++ 解法, 执行用时: 5ms, 内存消耗: 772KB, 提交时间: 2022-07-24

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param s string字符串 
     * @return int整型vector
     */
    vector<int> splitString(string s) {
        // write code here
      int n = s.size();
      typedef pair<int,int> PII;
      vector<PII> mp(26);
      for (int i = 1; i <= n; i ++)
      {
        int idx = s[i - 1] - 'a';
        if(mp[idx].first == 0)  mp[idx].first = i;
        mp[idx].second = i;
      }
      sort(mp.begin(), mp.end());
      //for(auto it : mp) cout << it.first << " " << it.second << endl;
      vector<int> res;
      vector<PII> merge;
      int start = 0, end = 0;
      for (int i = 0; i < 26; i ++)
      {
        auto x = mp[i].first, y = mp[i].second;
        if (x == 0) continue;
        if (start == 0) start = x, end = y;
        if(x <= end) end = max(end, y);
        else
        {
          merge.push_back({start, end});
          start = x, end = y;
        }
      }
      
      merge.push_back({start, end});
      cout << endl;
      //for(auto it : merge) cout << it.first << " " << it.second << endl;
      int m = merge.size();
      for (int i = 0; i < m; i ++)
      {
        auto x = merge[i].first, y = merge[i].second;
        //cout << x << "  xxxx  " << y << endl;
        res.push_back((y - x + 1));
      }
      return res;
    }
};

C++ 解法, 执行用时: 5ms, 内存消耗: 776KB, 提交时间: 2022-07-03

class Solution {
  public:
    vector<int> splitString(string s) {
        vector<int> v(26);
        for (int i = 0; i < s.size(); ++i) {
            v[s[i] - 'a'] = i;
        }
        vector<int> res;
        int z = 0;
        int a = 0;
        for (int i = 0; i < s.size(); ++i) {
            z = max(z, v[s[i] - 'a']);
            if (i == z) {
                res.emplace_back(z - a + 1);
                a = i + 1;
            }
        }
        return res;
    }
};