- DI String MatchGiven a string S that only contains “I” (increase) or “D” (decrease), let N = S.length.Return any permutation A of [0, 1, …, N] such that for all i = 0, …, N-1:If S[i] == “I”, then A[i] < A[i+1]If S[i] == “D”, then A[i] > A[i+1]Example 1:Input: “IDID"Output: [0,4,1,3,2]Example 2:Input: “III"Output: [0,1,2,3]Example 3:Input: “DDI"Output: [3,2,0,1]Note:1 <= S.length <= 10000S only contains characters “I” or “D”.题目地址算法：默认是递增的，然后根据D修改顺序，每遇到一串D，就reverse一段A。java代码class Solution { public int[] diStringMatch(String S) { int[] A = new int[S.length()+1]; for(int i=0;i<=S.length();i++){ A[i] = i; } char[] c = S.toCharArray(); int count=0; for(int i=0;i<c.length;i++){ if(c[i]==‘I’){ continue; } while(i<c.length && c[i]==‘D’){ i++; count++; } reverse(A, i-count, i); count=0; } return A; } private void reverse(int[] A, int begin, int end){ for(int i=0;i<(end-begin+1)/2;i++){ int temp=A[begin+i]; A[begin+i] = A[end-i]; A[end-i] = temp; } }}