golang 解法, 执行用时: 215 ms, 内存消耗: 22.2 MB, 提交时间: 2024-05-26 19:09:42

func queryResults(_ int, queries [][]int) []int {
	ans := make([]int, len(queries))
	color := map[int]int{}
	cnt := map[int]int{}
	for i, q := range queries {
		x, y := q[0], q[1]
		if c := color[x]; c > 0 {
			cnt[c]--
			if cnt[c] == 0 {
				delete(cnt, c)
			}
		}
		color[x] = y
		cnt[y]++
		ans[i] = len(cnt)
	}
	return ans
}

cpp 解法, 执行用时: 401 ms, 内存消耗: 153.5 MB, 提交时间: 2024-05-26 19:09:28

class Solution {
public:
    vector<int> queryResults(int, vector<vector<int>>& queries) {
        vector<int> ans;
        unordered_map<int, int> color, cnt;
        for (auto& q : queries) {
            int x = q[0], y = q[1];
            if (auto it = color.find(x); it != color.end()) {
                int c = it->second;
                if (--cnt[c] == 0) {
                    cnt.erase(c);
                }
            }
            color[x] = y;
            cnt[y]++;
            ans.push_back(cnt.size());
        }
        return ans;
    }
};

java 解法, 执行用时: 43 ms, 内存消耗: 80.2 MB, 提交时间: 2024-05-26 19:09:12

public class Solution {
    public int[] queryResults(int limit, int[][] queries) {
        int[] ans = new int[queries.length];
        Map<Integer, Integer> color = new HashMap<>();
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int i = 0; i < queries.length; i++) {
            int[] q = queries[i];
            int x = q[0];
            int y = q[1];
            if (color.containsKey(x)) {
                int c = color.get(x);
                if (cnt.merge(c, -1, Integer::sum) == 0) {
                    cnt.remove(c);
                }
            }
            color.put(x, y);
            cnt.merge(y, 1, Integer::sum);
            ans[i] = cnt.size();
        }
        return ans;
    }
}

python3 解法, 执行用时: 168 ms, 内存消耗: 60.9 MB, 提交时间: 2024-05-26 19:08:58

'''
用哈希表 color 维护第 x 个球的颜色，哈希表 cnt 维护每种颜色的出现次数。
'''
class Solution:
    def queryResults(self, _: int, queries: List[List[int]]) -> List[int]:
        ans = []
        color = {}
        cnt = defaultdict(int)
        for x, y in queries:
            if x in color:
                c = color[x]
                cnt[c] -= 1
                if cnt[c] == 0:
                    del cnt[c]
            color[x] = y
            cnt[y] += 1
            ans.append(len(cnt))
        return ans