p399

发表于 | 分类于 |

LeetCode p399 Kth Largest Element in an Array 题解

1.题目：

Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.
Example:
Given a / b = 2.0, b / c = 3.0.
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .
return [6.0, 0.5, -1.0, 1.0, -1.0 ].
The input is: vector<pair<string, string>> equations, vector& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector.
According to the example above:
equations = [ [“a”, “b”], [“b”, “c”] ],
values = [2.0, 3.0],
queries = [ [“a”, “c”], [“b”, “a”], [“a”, “e”], [“a”, “a”], [“x”, “x”] ].
The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

题意：

给你一些已知的换算法则，推断要求的表达式的值。如果无法推断出，则值为-1。

2.解题思路:

使用dfs进行 搜索。
首先是数据的储存，HashMap<String, HashMap<String, Double>> 每个出现过的字母记录下自己作为分子时，对应不同出现过的分母可以得到的值。

3.代码

[title] [] [url] [link text]
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
 
public class Solution {
    public double[] calcEquation(String[][] equations, double[] values, String[][] queries) {
        double[] ans = new double[queries.length];
		HashMap<String, HashMap<String, Double>> map = new HashMap<String, HashMap<String, Double>>();
		for (int i = 0; i < equations.length; i++) {
			if (!map.containsKey(equations[i][0])) {
				HashMap<String, Double> hashMap = new HashMap<String, Double>();
				map.put(equations[i][0], hashMap);
			}
			if (!map.containsKey(equations[i][1])) {
				HashMap<String, Double> hashMap = new HashMap<String, Double>();
				map.put(equations[i][1], hashMap);
			}
			HashMap<String, Double> h0 = map.get(equations[i][0]);
			h0.put(equations[i][1], values[i]);
			map.put(equations[i][0], h0);

			HashMap<String, Double> h1 = map.get(equations[i][1]);
			h1.put(equations[i][0], 1 / values[i]);
			map.put(equations[i][1], h1);
		}
		for (int i = 0; i < ans.length; i++) {

			ans[i] = dfs(map, queries[i][0], queries[i][1]);
		}
		return ans;
    }
    public double dfs(HashMap<String, HashMap<String, Double>> map,
			String start, String end) {
		if (!map.containsKey(start))
			return -1;
		if (start.equals(end))
			return 1;
		if (map.get(start).containsKey(end)) {
			return map.get(start).get(end);
		}
		HashMap<String, Double> hashMap = map.remove(start);
		double a = -1;
		for (String s : hashMap.keySet()) {
			double b2 = dfs(map, s, end); // start/str=b1,str/end=b2;
			if (b2 != -1) {
				double b1 = hashMap.get(s);
				a = b1 * b2;
				break;
			}
		}
		map.put(start, hashMap); // 将这个map重新装回去
		return a;

	}
}

4.一些总结：