LeetCode | 767. Reorganize String调整字符串使得字符串相同字符不相邻

Givena string S, check if the letters can berearranged so that two characters that are adjacent
to each other are not thesame.
Ifpossible, output any possible result.  If not possible, return the emptystring.
Example 1:

Input: S = "aab"

Output: "aba"

Example 2:

Input: S = "aaab"

Output: ""

Note:
S will consist of lowercase letters and have length in range [1, 500].
这一题明明知道怎么做，但是由于思路不清晰，改了好多次，很烦

这一题首先需要明白，可以证明，当s中每一个字母的个数都小于等于其他字母个数之和减一的时候，一定存在一个方案可以使得字符串S的每一个字母不相邻，方法是，使用一个数组theLetter存储s中每一个字母的出现的次数，然后每次取theLetter中的前两个不为0的位置的值，将里面的字母交替放入result，然后取theLetter后两个值，交替放入result。。。一直持续这个过程，直到theLetter中只剩下一个不为0的组，然后把这个组的字母按照从前往后的顺序交替放入S的前面，假如能够放入而不产生冲突，那么S一定存在一种方案，如果产生冲突，那么S一定有一组的个数大于其他所有组的个数之和加一，那么一定不存在可行的方案

当对于某个问题想解决方法的时候，当有一种方法，当产生结果A的时候一定有解，当产生结果B的时候解一定不存在，那么这种方法可以直接被使用，不需要去证明！

class Solution {

public:

stringreorganizeString(string S) {

int theLetter[26] = { 0 };

string result = "";

int theLastI = 0; int theLNeed = 0; chartheNeedC;

for (int i = 0; i < S.size(); i++)

{

theLetter[S[i] - 'a']++;

}

for (int i = 0; i < 26; i++)

if (theLetter[i] != 0) theLastI = i;

int indexOne=0, indexTwo=0;

while(indexOne<26&&theLetter[indexOne] == 0) indexOne++;

indexTwo = indexOne; indexTwo++;

while (indexTwo < 26 &&theLetter[indexTwo] == 0) indexTwo++;

while (indexTwo < 26)

{

if (indexTwo == theLastI)

{

if (theLetter[indexOne] >theLetter[indexTwo])

{

theLNeed =theLetter[indexOne] - theLetter[indexTwo]-1;

theNeedC = indexOne +'a';

if (theLNeed >result.size()) return "";

}

else

{

theLNeed =theLetter[indexTwo] - theLetter[indexOne];

theNeedC = indexTwo +'a';

if (theLNeed >result.size()+1) return "";

}

}

while (1)

{

result.push_back(indexOne +'a'); theLetter[indexOne]--;

if (theLetter[indexOne] == 0)break;

result.push_back(indexTwo +'a');

theLetter[indexTwo]--;

if (theLetter[indexTwo] == 0) break;

}

if(theLetter[indexOne] != 0)

{

if(indexTwo==theLastI)result.push_back(indexOne + 'a');

while (indexTwo < 26&& theLetter[indexTwo] == 0) indexTwo++;

}

else

{

if (indexTwo == theLastI)result.push_back(indexTwo + 'a');

indexOne = indexTwo++;

while (indexTwo < 26&& theLetter[indexTwo] == 0) indexTwo++;

}

}

indexOne = 0;

for (int i = 0; i < theLNeed; i++)

{

result.insert(result.begin() +indexOne, theNeedC);

indexOne += 2;

}

return result;

}

};