LeetCode 942. 增減字串匹配(C、C++、python)
阿新 • • 發佈:2018-12-02
給定只含 "I"
(增大)或 "D"
(減小)的字串 S
,令 N = S.length
。
返回 [0, 1, ..., N]
的任意排列 A
使得對於所有 i = 0, ..., N-1
,都有:
- 如果
S[i] == "I"
,那麼A[i] < A[i+1]
- 如果
S[i] == "D"
,那麼A[i] > A[i+1]
示例 1:
輸出:"IDID"
輸出:[0,4,1,3,2]
示例 2:
輸出:"III"
輸出:[0,1,2,3]
示例 3:
輸出:"DDI"
輸出:[3,2,0,1]
提示:
1 <= S.length <= 1000
S
只包含字元 "I"
或 "D"
。
C
/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */ int* diStringMatch(char* S, int* returnSize) { int N=strlen(S); int* res=(int*)malloc(sizeof(int)*(N+1)); int a=N; int b=0; for(int i=0;i<N;i++) { if('I'==S[i]) { res[i]=b; b++; } else { res[i]=a; a--; } } res[N]=b; *returnSize=N+1; return res; }
C++
class Solution { public: vector<int> diStringMatch(string S) { int N=S.length(); int a=N; int b=0; vector<int> res(N+1,0); for(int i=0;i<N;i++) { if('I'==S[i]) { res[i]=b; b++; } else { res[i]=a; a--; } } res[N]=a; return res; } };
python
class Solution:
def diStringMatch(self, S):
"""
:type S: str
:rtype: List[int]
"""
N=len(S)
res=[0 for i in range(N+1)]
a=N
b=0
for i in range(N):
if 'I'==S[i]:
res[i]=b
b+=1
else:
res[i]=a
a-=1
res[N]=a
return res