N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A swap consists of choosing any two people, then they stand up and switch seats.
The people and seats are represented by an integer from
0
to 2N-1
, the couples are numbered in order, the first couple being (0, 1)
, the second couple being (2, 3)
, and so on with the last couple being (2N-2, 2N-1)
.
The couples' initial seating is given by
row[i]
being the value of the person who is initially sitting in the i-th seat.
Example 1:
Input: row = [0, 2, 1, 3] Output: 1 Explanation: We only need to swap the second (row[1]) and third (row[2]) person.
Example 2:
Input: row = [3, 2, 0, 1] Output: 0 Explanation: All couples are already seated side by side.
Note:
len(row)
is even and in the range of[4, 60]
.row
is guaranteed to be a permutation of0...len(row)-1
.
贪心,本质上Cycle Finding或union find
ref: https://leetcode.com/problems/couples-holding-hands/discuss/113362/JavaC++-O(N)-solution-using-cyclic-swapping
class Solution { public int minSwapsCouples(int[] row) { Map<Integer, Integer> valueToIndex = new HashMap<>(); for (int i = 0; i < row.length; i++) { valueToIndex.put(row[i], i); } int swaps = 0; for (int i = 0; i < row.length; i += 2) { int p1 = row[i]; int p2 = (p1 % 2) == 0 ? p1 + 1 : p1 - 1; if (row[i + 1] != p2) { int p2Index = valueToIndex.get(p2); valueToIndex.put(p2, i + 1); valueToIndex.put(row[i + 1], p2Index); swap(row, p2Index, i + 1); swaps++; } } return swaps; } private void swap(int[] row, int i1, int i2) { int tmp = row[i1]; row[i1] = row[i2]; row[i2] = tmp; } }
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