题目描述:
给定只含 "I"
(增大)或 "D"
(减小)的字符串 S
,令 N = S.length
。
返回 [0, 1, ..., N]
的任意排列 A
使得对于所有 i = 0, ..., N-1
,都有:
如果 S[i] == "I"
,那么 A[i] < A[i+1]
如果 S[i] == "D"
,那么 A[i] > A[i+1]
示例 1:
输出:"IDID"
输出:[0,4,1,3,2]
示例 2:
输出:"III"
输出:[0,1,2,3]
示例 3:
输出:"DDI"
输出:[3,2,0,1]
提示:
1 <= S.length <= 1000
S
只包含字符"I"
或"D"
。
解法:
class Solution {
public:
vector<int> diStringMatch(string S) {
//ID
//0 2 1
//1 2 0
//II
//0 1 2
//DD
//2 1 0
//DID
//3 0 2 1
// greedy algorithm
int l = 0, r = S.size();
vector<int> res;
for(char ch : S){
if(ch == 'I'){
res.push_back(l++);
}else{
res.push_back(r--);
}
}
res.push_back(l);
return res;
}
};