[Algo] 73. Combinations Of Coins

Given a number of different denominations of coins (e.g., 1 cent, 5 cents, 10 cents, 25 cents), get all the possible ways to pay a target number of cents.

Arguments

• coins - an array of positive integers representing the different denominations of coins, there are no duplicate numbers and the numbers are sorted by descending order, eg. {25, 10, 5, 2, 1}
• target - a non-negative integer representing the target number of cents, eg. 99

Assumptions

• coins is not null and is not empty, all the numbers in coins are positive
• target >= 0
• You have infinite number of coins for each of the denominations, you can pick any number of the coins.

Return

• a list of ways of combinations of coins to sum up to be target.
• each way of combinations is represented by list of integer, the number at each index means the number of coins used for the denomination at corresponding index.

Examples

coins = {2, 1}, target = 4, the return should be

[

  [0, 4],   (4 cents can be conducted by 0 * 2 cents + 4 * 1 cents)

  [1, 2],   (4 cents can be conducted by 1 * 2 cents + 2 * 1 cents)

  [2, 0]    (4 cents can be conducted by 2 * 2 cents + 0 * 1 cents)

]

public class Solution {
  public List<List<Integer>> combinations(int target, int[] coins) {
    // Write your solution here
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> list = new ArrayList<>();
    helper(res, list, 0, coins, target);
    return res;
  }

    private void helper(List<List<Integer>> res, List<Integer> list, int index, int[] coins, int left) {
    if (index == coins.length) {
        if (left == 0) {
          res.add(new ArrayList<>(list));
      }
      return;
    }
    for (int i = 0; i <= left/coins[index]; i++) {
      list.add(i);
      helper(res, list, index + 1, coins, left - i * coins[index]);
      list.remove(list.size() - 1);
    }
  }
}

原文：https://www.cnblogs.com/xuanlu/p/12316173.html