By now, you are given a secret signature consisting of character 'D' and 'I'. 'D' represents a decreasing relationship between two numbers, 'I' represents an increasing relationship between two numbers. And our secret signature was constructed by a special integer array, which contains uniquely all the different number from 1 to n (n is the length of the secret signature plus 1). For example, the secret signature "DI" can be constructed by array [2,1,3] or [3,1,2], but won't be constructed by array [3,2,4] or [2,1,3,4], which are both illegal constructing special string that can't represent the "DI" secret signature.On the other hand, now your job is to find the lexicographically smallest permutation of [1, 2, ... n] could refer to the given secret signature in the input.Example 1:Input: "I"Output: [1,2]Explanation: [1,2] is the only legal initial spectial string can construct secret signature "I", where the number 1 and 2 construct an increasing relationship.Example 2:Input: "DI"Output: [2,1,3]Explanation: Both [2,1,3] and [3,1,2] can construct the secret signature "DI", but since we want to find the one with the smallest lexicographical permutation, you need to output [2,1,3]Note:The input string will only contain the character 'D' and 'I'.The length of input string is a positive integer and will not exceed 10,000
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For example, given IDIIDD we start with sorted sequence 1234567
k continuous D starting at index i we need to reverse [i, i+k] portion of the sorted sequence. IDIIDD1234567 // sorted1324765 // answer
1 public class Solution { 2 public int[] findPermutation(String s) { 3 int n = s.length(), arr[] = new int[n + 1]; 4 for (int i = 0; i <= n; i++) arr[i] = i + 1; // sorted 5 for (int h = 0; h < n; h++) { 6 if (s.charAt(h) == 'D') { 7 int l = h; 8 while (h < n && s.charAt(h) == 'D') h++; 9 reverse(arr, l, h); 10 } 11 } 12 return arr;13 } 14 15 void reverse(int[] arr, int l, int h) {16 while (l < h) {17 arr[l] ^= arr[h];18 arr[h] ^= arr[l];19 arr[l] ^= arr[h];20 l++; h--;21 } 22 }23 }