[LeetCode] 846. Hand of Straights
Alice has some number of cards and she wants to rearrange the cards into groups so that each group is of size groupSize
, and consists of groupSize
consecutive cards.
Given an integer array hand
where hand[i]
is the value written on the ith
card and an integer groupSize
, return true
if she can rearrange the cards, or false
otherwise.
Example 1:
Input: hand = [1,2,3,6,2,3,4,7,8], groupSize = 3 Output: true Explanation: Alice's hand can be rearranged as [1,2,3],[2,3,4],[6,7,8]
Example 2:
Input: hand = [1,2,3,4,5], groupSize = 4 Output: false Explanation: Alice's hand can not be rearranged into groups of 4
Constraints:
1 <= hand.length <= 104
0 <= hand[i] <= 109
1 <= groupSize <= hand.length
Note: This question is the same as 1296: https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/
一手顺子。
Alice 手中有一把牌,她想要重新排列这些牌,分成若干组,使每一组的牌数都是 groupSize ,并且由 groupSize 张连续的牌组成。
给你一个整数数组 hand 其中 hand[i] 是写在第 i 张牌上的数值。如果她可能重新排列这些牌,返回 true ;否则,返回 false 。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/hand-of-straights
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这个题的思路别想复杂了。首先需要对数组排序,同时需要对数组的所有元素用hashmap进行统计。比如第一个例子,排序之后长这样,
[1,2,3,6,2,3,4,7,8] ->[1,2,2,3,3,4,6,7,8]
然后对于每一个数字 N,我们要找的是是否存在数字 N + 1,同时 N + 1 出现的次数仍然大于 0,如果不满足则直接返回 false。如果满足,则接着去检查 N + 2, N + 3,直到 N + W。
时间O(nlogn)
空间O(n)
Java实现
1 class Solution { 2 public boolean isNStraightHand(int[] hand, int W) { 3 // corner case 4 if (hand.length % W != 0) { 5 return false; 6 } 7 8 // normal case 9 HashMap<Integer, Integer> map = new HashMap<>(); 10 for (int num : hand) { 11 map.put(num, map.getOrDefault(num, 0) + 1); 12 } 13 Arrays.sort(hand); 14 for (int i = 0; i < hand.length; i++) { 15 if (map.get(hand[i]) == 0) { 16 continue; 17 } 18 int cur = hand[i]; 19 int end = cur + W; 20 while (cur < end) { 21 int count = map.getOrDefault(cur, 0); 22 if (count == 0) { 23 return false; 24 } 25 map.put(cur, count - 1); 26 cur++; 27 } 28 } 29 return true; 30 } 31 }
JavaScript实现
1 /** 2 * @param {number[]} hand 3 * @param {number} W 4 * @return {boolean} 5 */ 6 var isNStraightHand = function (hand, W) { 7 if (hand.length % W !== 0) { 8 return false; 9 } 10 11 let cards = {}; 12 hand.forEach((card) => { 13 cards[card] ? cards[card]++ : (cards[card] = 1); 14 }); 15 16 for (let i = 0; i < hand.length; i++) { 17 if (cards[hand[i] - 1] > 0 || !cards[hand[i]]) { 18 continue; 19 } 20 let min = hand[i]; 21 let j = 0; 22 while (j < W) { 23 if (cards[min] > 0) { 24 cards[min]--; 25 min++; 26 } else { 27 return false; 28 } 29 j++; 30 } 31 } 32 return true; 33 };
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