leetcode399. Evaluate Division

摘要： 題目要求 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 numb...

題目要求

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<double>& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector<double>.

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.

已知一些字母之間的關係式，問是否能夠計算出其它字母之間的倍數關係？

如已知a/b=2.0 b/c=3.0 問是否能夠計算出a/c, b/a, a/e, a/a, x/x 的值。如果無法計算得出，則返回-1。這裡x/x 的值因為在條件中無法獲知x是否等於零，因此也無法計算其真實結果，也需要返回-1。

思路和程式碼

假如我們將除數和被除數看做是圖的頂點，將除數和被除數之間的倍數關係試做二者之間邊的權重。即a/b=2.0 代表點a指向點b的邊的權重為2.0，而點b指向點a的邊的全中為1/2.0=0.5。

因此我們可以將輸入的表示式轉化為一個加權有向圖，而題目的問題則被轉化為求兩個點之間是否能夠找到一條邊，如果無法找到，則返回-1，否則返回路徑上每條邊的權重的乘積。

程式碼如下：

public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
//圖的鏈式表示法
Map<String, List<String>> pairs = new HashMap<>();
//圖上每條邊的權重
Map<String, List<Double>> valuedPairs = new HashMap<>();
for(int i = 0 ; i < equations.size() ; i++) {
//獲取第i個方程式
List<String> equation = equations.get(i);
String multiplied = equation.get(0);//被除數
String multiplier = equation.get(1);//除數
//如果被除數從來沒有新增到圖中，則將其作為頂點在圖中初始化
if(!pairs.containsKey(multiplied)) {
pairs.put(multiplied, new ArrayList<>());
valuedPairs.put(multiplied, new ArrayList<>());
}
//如果除數從來沒有新增到圖中，則將其作為頂點在圖中初始化
if(!pairs.containsKey(multiplier)) {
pairs.put(multiplier, new ArrayList<>());
valuedPairs.put(multiplier, new ArrayList<>());
}
//新增邊和邊的權重
pairs.get(multiplied).add(multiplier);
pairs.get(multiplier).add(multiplied);
valuedPairs.get(multiplied).add(values[i]);
valuedPairs.get(multiplier).add(1.0 / values[i]);
}

//結果集
double[] result = new double[queries.size()];
for(int i = 0 ; i<queries.size() ; i++) {
//在圖中，以被除數作為頂點，深度優先遍歷圖，直到找到值為除數的頂點
result[i] = dfs(queries.get(i).get(0), queries.get(i).get(1), pairs, valuedPairs, new HashSet<>(), 1.0);
result[i] = result[i]==0.0 ? -1.0 : result[i];
}
return result;
}

public double dfs(String multiplied, String multiplier, Map<String, List<String>> pairs, Map<String, List<Double>> valuedPairs, Set<String> visited, double curResult) {
//如果圖中不包含該被除數頂點，則無法獲知該表示式的值
if(!pairs.containsKey(multiplied)) return 0.0;
//如果再次訪問過該被除數，說明找到了一條環路，則該深度優先遍歷結果失敗，直接拋棄
if(visited.contains(multiplied)) return 0.0;
//如果被除數等於除數，則返回1.0
if(multiplied.equals(multiplier)) return curResult;
visited.add(multiplied);
//獲得當前被除數的所有鄰接頂點
List<String> multipliers = pairs.get(multiplied);
//獲得所有鄰接邊的權重
List<Double> multiplierValues = valuedPairs.get(multiplied);
double tmp = 0.0;
for(int i = 0 ; i<multipliers.size() ; i++) {
//以鄰接點為新的頂點，繼續深度優先遍歷
//此時呼叫方法中curResult的值代表的是該原鄰接點除以鄰接點的值
//如 a/b=2, b/c=3, 則a=2b，因此當我們以b作為鄰接點尋找c時，需要記錄原被除數是現被除數的兩倍
tmp = dfs(multipliers.get(i), multiplier, pairs, valuedPairs, visited, curResult * multiplierValues.get(i));
//找到非零路徑，結束深度優先遍歷
if(tmp != 0.0){
break;
}
}
visited.remove(multiplied);
return tmp;
}