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:
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Input: "I"
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Output:
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[1,2]
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Explanation:
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[1,2] is the only legal initial spectial string can construct secret signature "I", where the number 1 and 2 construct an increasing relationship.
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Example 2:
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Input: "DI"
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Output: [2,1,3]
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Explanation:
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Both [2,1,3] and [3,1,2] can construct the secret signature "DI",
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but since we want to find the one with the smallest lexicographical permutation, you need to output [2,1,3]
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# 2. Implementation

(1) Two Pointers

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class Solution {
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public int[] findPermutation(String s) {
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int n = s.length();
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int[] res = new int[n + 1];
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for (int i = 0; i <= n; i++) {
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res[i] = i + 1;
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}
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int i = 0;
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while (i < n) {
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if (s.charAt(i) == 'D') {
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int j = i;
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while (j < n && s.charAt(j) == 'D') {
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++j;
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}
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reverse(res, i, j);
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i = j;
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}
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++i;
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}
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return res;
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}
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public void reverse(int[] nums, int i, int j) {
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while (i < j) {
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int temp = nums[i];
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nums[i] = nums[j];
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nums[j] = temp;
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++i;
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--j;
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}
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}
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}
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(2) Stack:

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