484 Find Permutation

484. Find Permutation

1. Question

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]

2. Implementation

(1) Two Pointers

思路:

假设 input String = "DDID"

   原数组为         1 2 3 4 5

   变换后得          3 2 1 5 4

通过观察例子我们可以看到，string的[0,1]上有两个D，而对应的数组[0, 2]的数字被reversed. 同样，string在[3]上有D，而nums对应的位置[3,4]的数字也reversed，所以当D在string[i, j], i <= j出现时， 对应的nums数组在[i, j + 1]的数字要reverse

class Solution {
    public int[] findPermutation(String s) {
        int n = s.length();

        int[] res = new int[n + 1];

        for (int i = 0; i <= n; i++) {
            res[i] = i + 1;
        }

        int i = 0;
        while (i < n) {
            if (s.charAt(i) == 'D') {
                int j = i;

                while (j < n && s.charAt(j) == 'D') {
                    ++j;
                }
                reverse(res, i, j);
                i = j;
            }
            ++i;
        }
        return res;
    }

    public void reverse(int[] nums, int i, int j) {
        while (i < j) {
            int temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
            ++i;
            --j;
        }
    }
}

(2) Stack:

思路和上一种做法类似，只不过我们是通过stack达到reverse的效果。注意在最后我们要在stack加上s.length() + 1这个数

class Solution {
    public int[] findPermutation(String s) {
        int n = s.length();
        int[] res = new int[n + 1];

        Stack<Integer> stack = new Stack<>();
        int j = 0;
        for (int i = 1; i <= n; i++) {
            if (s.charAt(i - 1) == 'D') {
                stack.push(i);
            }
            else {
                stack.push(i);
                while (!stack.isEmpty()) {
                    res[j] = stack.pop();
                    ++j;
                }
            }
        }
        stack.push(n + 1);
        while (!stack.isEmpty()) {
            res[j] = stack.pop();
            ++j;
        }
        return res;
    }
}

3. Time & Space Complexity

Two Pointers: 时间复杂度O(n), 空间复杂度O(1)

Stack: 时间复杂度O(n), 空间复杂度O(n)