Q942 增减字符串匹配
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"
。
class Solution942 {
public int[] diStringMatch(String S) {
int[] nums = new int[S.length() + 1];
char[] chs = S.toCharArray();
int n = 0;
for (char c : chs) {
if (c == 'D')
n++;
}
int m = n + 1;
nums[0] = n--;
for (int i = 0; i < chs.length; i++) {
if (chs[i] == 'I')
nums[i + 1] = m++;
else
nums[i + 1] = n--;
}
return nums;
}
}