Merge Similar Items
You are given two 2D integer arrays, items1 and items2, representing two sets of items. Each array items has the following properties:
items[i] = [valuei, weighti] where valuei represents the value and weighti represents the weight of the ith item.
The value of each item in items is unique.
Return a 2D integer array ret where ret[i] = [valuei, weighti], with weighti being the sum of weights of all items with value valuei.
Note: ret should be returned in ascending order by value.
Example 1:
Input: items1 = [[1,1],[4,5],[3,8]], items2 = [[3,1],[1,5]]
Output: [[1,6],[3,9],[4,5]]
Explanation:
The item with value = 1 occurs in items1 with weight = 1 and in items2 with weight = 5, total weight = 1 + 5 = 6.
The item with value = 3 occurs in items1 with weight = 8 and in items2 with weight = 1, total weight = 8 + 1 = 9.
The item with value = 4 occurs in items1 with weight = 5, total weight = 5.
Therefore, we return [[1,6],[3,9],[4,5]].
Example 2:
Input: items1 = [[1,1],[3,2],[2,3]], items2 = [[2,1],[3,2],[1,3]]
Output: [[1,4],[2,4],[3,4]]
Explanation:
The item with value = 1 occurs in items1 with weight = 1 and in items2 with weight = 3, total weight = 1 + 3 = 4.
The item with value = 2 occurs in items1 with weight = 3 and in items2 with weight = 1, total weight = 3 + 1 = 4.
The item with value = 3 occurs in items1 with weight = 2 and in items2 with weight = 2, total weight = 2 + 2 = 4.
Therefore, we return [[1,4],[2,4],[3,4]].
Example 3:
Input: items1 = [[1,3],[2,2]], items2 = [[7,1],[2,2],[1,4]]
Output: [[1,7],[2,4],[7,1]]
Explanation:
The item with value = 1 occurs in items1 with weight = 3 and in items2 with weight = 4, total weight = 3 + 4 = 7.
The item with value = 2 occurs in items1 with weight = 2 and in items2 with weight = 2, total weight = 2 + 2 = 4.
The item with value = 7 occurs in items2 with weight = 1, total weight = 1.
Therefore, we return [[1,7],[2,4],[7,1]].
Constraints:
1 <= items1.length, items2.length <= 1000
items1[i].length == items2[i].length == 2
1 <= valuei, weighti <= 1000
Each valuei in items1 is unique.
Each valuei in items2 is unique.
思路一:用 TreeSet 存储 key,用 map 存储二维数组,最后遍历 TreeSet。看了一下评论区,发现直接用 TreeMap 就可以实现存储有序的值,可以省去 TreeSet
public List<List<Integer>> mergeSimilarItems(int[][] items1, int[][] items2) {
List<List<Integer>> result = new ArrayList<>();
Set<Integer> keys = new TreeSet<>();
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < items1.length; i++) {
keys.add(items1[i][0]);
int finalI = i;
map.compute(items1[i][0], (k, v) -> v == null ? items1[finalI][1] : v + items1[finalI][1]);
}
for (int i = 0; i < items2.length; i++) {
int finalI = i;
keys.add(items2[i][0]);
map.compute(items2[i][0], (k, v) -> v == null ? items2[finalI][1] : v + items2[finalI][1]);
}
for (Integer key : keys) {
List<Integer> list = new ArrayList<>();
list.add(key);
list.add(map.get(key));
result.add(list);
}
return result;
}
标签:weight,value,occurs,items2,item,items1,easy,2363,leetcode
From: https://www.cnblogs.com/iyiluo/p/16867224.html