C++ 实现字典排序以查找唯一排列

我试图实现一种算法，以生成所描述的字典顺序的唯一排列。下面是我的实现

void My_Permute(string word) {
    sort(word.begin(),word.end());
    int size = word.size(); 
    while (true) {
        cout << word << endl;
        int i = size - 2;  
        for (; i >= 0; --i) { 
            if (word[i] < word[i+1]) break; 
        } 
        if (i <= -1) break; 
        swap(word[i],word[i+1]);
        cout << word << endl;
        reverse(word.begin()+i+1,word.end());
    } 
} 

链接算法缺少第2步

void Permute(string word) {
    sort(word.begin(), word.end());
    int size = word.size();
    while (true) {
        cout << word << endl;
        // Find the largest index k such that a[k] < a[k + 1].
        // If no such index exists, the permutation is the last permutation.
        int i = size - 2;
        for (; i >= 0; --i) {
            if (word[i] < word[i+1])
                break;
        }
        if (i < 0)
            break;
        // Find the largest index l such that a[k] < a[l].
        // Since k + 1 is such an index, l is well defined and satisfies k < l.
        int j = size - 1;
        for (; ; --j) {
            if (word[i] < word[j])
                break;
        }
        // Swap a[k] with a[l].
        swap(word[i], word[j]);
        // Reverse the sequence from a[k + 1] up to and including the
        // final element a[n].
        reverse(word.begin() + i + 1, word.end());
    }
}

24（即4！）行，如预期

---

顺便说一句，你应该尽可能多地使用它。

风格改进建议：不要使用

if（讽刺的是，你在解决方案中使用
使用namespace std
it；）@理想化只是为了避免重新格式化代码的复杂性，这样版本之间的差异就会更清楚：但是我在没有itI的情况下测试了它，就像你使用的网站一样。谢谢你的知识。
void Permute(string word) {
    sort(word.begin(), word.end());
    int size = word.size();
    while (true) {
        cout << word << endl;
        // Find the largest index k such that a[k] < a[k + 1].
        // If no such index exists, the permutation is the last permutation.
        int i = size - 2;
        for (; i >= 0; --i) {
            if (word[i] < word[i+1])
                break;
        }
        if (i < 0)
            break;
        // Find the largest index l such that a[k] < a[l].
        // Since k + 1 is such an index, l is well defined and satisfies k < l.
        int j = size - 1;
        for (; ; --j) {
            if (word[i] < word[j])
                break;
        }
        // Swap a[k] with a[l].
        swap(word[i], word[j]);
        // Reverse the sequence from a[k + 1] up to and including the
        // final element a[n].
        reverse(word.begin() + i + 1, word.end());
    }
}

abcd
abdc
acbd
acdb
adbc
adcb
bacd
badc
bcad
bcda
bdac
bdca
cabd
cadb
cbad
cbda
cdab
cdba
dabc
dacb
dbac
dbca
dcab
dcba