Google – Minimum Fluctuation Sequence


Google – Minimum Fluctuation Sequence
给一个String pattern, 包好两个字母i和d,i表示increase, d表示decrease, 生成一个最小的且满足pattern的数字,比如:
iii -> 1234
ddd -> 4321
iid -> 1243
每个数字不能重复使用。能重复使用的解法见后面follow up.
[Solution]
Greedy + Linked list
维护一个linked list来存每个digit,用一个pointer来指向最近一次碰到d的digit,一旦碰到i,pointer就指向tail.
[Note]
需要注意一下0这个特殊情况,因为第一个digit不能为0,所以pattern中第一个d肯定对应0。需要特殊处理一下。

class Solution {
  public int fluctuationSequence(String pattern) {
    if (pattern == null || pattern.isEmpty()) {
      return 0;
    }
     
    List<Integer> list = new LinkedList<>();
    list.add(1);
    int next = 2;
    int idx = 0;
    boolean firstD = true;
    for (char c : pattern.toCharArray()) {
      if (c == 'i') {
        list.add(next);
        idx = list.size() - 1;
      }
      else if (firstD) {
        list.add(0);
        firstD = false;
        continue;
      }
      else {
        list.add(idx, next);
      }
      next++;
    }
     
    int result = 0;
    for (int num : list) {
      result = result * 10 + num;
    }
    return result;
  }
}
 
// digit能重复, candy解法
class Solution2 {
  public int fluctuationSequence(String pattern) {
    if (pattern == null || pattern.isEmpty()) {
      return 0;
    }
     
    int[] nums = new int[pattern.length() + 1];
    nums[0] = 1;
    int prev = 1;
    boolean firstD = true;
    for (int i = 0; i < pattern.length(); i++) {
      char c = pattern.charAt(i);
      if (c == 'i') {
        nums[i + 1] = prev + 1;
        prev = prev + 1;
      }
      else if (firstD) {
        nums[i + 1] = 0;
        prev = 0;
        firstD = false;
      }
      else if (prev == 0) {
        nums[i + 1] = 0;
      }
      else {
        nums[i + 1] = prev - 1;
        prev = prev - 1;
      }
    }
     
    for (int i = pattern.length() - 1; i >= 0; i--) {
      char c = pattern.charAt(i);
      if (c == 'i' && nums[i] >= nums[i + 1]) {
        nums[i + 1] = nums[i] + 1;
      }
      else if (c == 'd' && nums[i] <= nums[i + 1]) {
        nums[i] = nums[i + 1] + 1;
      }
    }
     
    int result = 0;
    for (int num : nums) {
      result = result * 10 + num;
    }
    return result;
  }
}

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