[LeetCode] 1899. Merge Triplets to Form Target Triplet
作者:互联网
A triplet is an array of three integers. You are given a 2D integer array triplets
, where triplets[i] = [ai, bi, ci]
describes the ith
triplet. You are also given an integer array target = [x, y, z]
that describes the triplet you want to obtain.
To obtain target
, you may apply the following operation on triplets
any number of times (possibly zero):
- Choose two indices (0-indexed)
i
andj
(i != j
) and updatetriplets[j]
to become[max(ai, aj), max(bi, bj), max(ci, cj)]
.- For example, if
triplets[i] = [2, 5, 3]
andtriplets[j] = [1, 7, 5]
,triplets[j]
will be updated to[max(2, 1), max(5, 7), max(3, 5)] = [2, 7, 5]
.
- For example, if
Return true
if it is possible to obtain the target
triplet [x, y, z]
as an element of triplets
, or false
otherwise.
Example 1:
Input: triplets = [[2,5,3],[1,8,4],[1,7,5]], target = [2,7,5] Output: true Explanation: Perform the following operations: - Choose the first and last triplets [[2,5,3],[1,8,4],[1,7,5]]. Update the last triplet to be [max(2,1), max(5,7), max(3,5)] = [2,7,5]. triplets = [[2,5,3],[1,8,4],[2,7,5]] The target triplet [2,7,5] is now an element of triplets.
Example 2:
Input: triplets = [[1,3,4],[2,5,8]], target = [2,5,8] Output: true Explanation: The target triplet [2,5,8] is already an element of triplets.
Example 3:
Input: triplets = [[2,5,3],[2,3,4],[1,2,5],[5,2,3]], target = [5,5,5] Output: true Explanation: Perform the following operations: - Choose the first and third triplets [[2,5,3],[2,3,4],[1,2,5],[5,2,3]]. Update the third triplet to be [max(2,1), max(5,2), max(3,5)] = [2,5,5]. triplets = [[2,5,3],[2,3,4],[2,5,5],[5,2,3]]. - Choose the third and fourth triplets [[2,5,3],[2,3,4],[2,5,5],[5,2,3]]. Update the fourth triplet to be [max(2,5), max(5,2), max(5,3)] = [5,5,5]. triplets = [[2,5,3],[2,3,4],[2,5,5],[5,5,5]]. The target triplet [5,5,5] is now an element of triplets.
Example 4:
Input: triplets = [[3,4,5],[4,5,6]], target = [3,2,5] Output: false Explanation: It is impossible to have [3,2,5] as an element because there is no 2 in any of the triplets.
Constraints:
1 <= triplets.length <= 105
triplets[i].length == target.length == 3
1 <= ai, bi, ci, x, y, z <= 1000
合并若干三元组以形成目标三元组。
三元组 是一个由三个整数组成的数组。给你一个二维整数数组 triplets ,其中 triplets[i] = [ai, bi, ci] 表示第 i 个 三元组 。同时,给你一个整数数组 target = [x, y, z] ,表示你想要得到的 三元组 。
为了得到 target ,你需要对 triplets 执行下面的操作 任意次(可能 零 次):
选出两个下标(下标 从 0 开始 计数)i 和 j(i != j),并 更新 triplets[j] 为 [max(ai, aj), max(bi, bj), max(ci, cj)] 。
例如,triplets[i] = [2, 5, 3] 且 triplets[j] = [1, 7, 5],triplets[j] 将会更新为 [max(2, 1), max(5, 7), max(3, 5)] = [2, 7, 5] 。
如果通过以上操作我们可以使得目标 三元组 target 成为 triplets 的一个 元素 ,则返回 true ;否则,返回 false 。来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/merge-triplets-to-form-target-triplet
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思路是贪心,具体做法就是遍历 triplets,然后三元组的每一元都取最大值,看看能否凑出最后的target。同时注意如果遍历的时候遇到某个 triplet 的任何一元大于 target 中对应的那一元的时候,这个 triplet 就可以跳过了。
时间O(n)
空间O(1)
Java实现
1 class Solution { 2 public boolean mergeTriplets(int[][] triplets, int[] target) { 3 int[] res = new int[3]; 4 for (int[] t : triplets) { 5 if (t[0] <= target[0] && t[1] <= target[1] && t[2] <= target[2]) { 6 res = new int[] { Math.max(res[0], t[0]), Math.max(res[1], t[1]), Math.max(res[2], t[2]) }; 7 } 8 } 9 return Arrays.equals(res, target); 10 } 11 }
标签:Triplet,target,Form,int,max,三元组,triplet,triplets,Target 来源: https://www.cnblogs.com/cnoodle/p/14883934.html