摘要：題目要求已知一些字母之間的關(guān)系式，問是否能夠計(jì)算出其它字母之間的倍數(shù)關(guān)系如已知問是否能夠計(jì)算出的值。這里的值因?yàn)樵跅l件中無法獲知是否等于零，因此也無法計(jì)算其真實(shí)結(jié)果，也需要返回。即代表點(diǎn)指向點(diǎn)的邊的權(quán)重為，而點(diǎn)指向點(diǎn)的邊的全中為。
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> equations, vector& values, vector> 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.
已知一些字母之間的關(guān)系式，問是否能夠計(jì)算出其它字母之間的倍數(shù)關(guān)系？
如已知a/b=2.0 b/c=3.0問是否能夠計(jì)算出a/c, b/a, a/e, a/a, x/x的值。如果無法計(jì)算得出，則返回-1。這里x/x的值因?yàn)樵跅l件中無法獲知x是否等于零，因此也無法計(jì)算其真實(shí)結(jié)果，也需要返回-1。
思路和代碼
假如我們將除數(shù)和被除數(shù)看做是圖的頂點(diǎn)，將除數(shù)和被除數(shù)之間的倍數(shù)關(guān)系試做二者之間邊的權(quán)重。即a/b=2.0代表點(diǎn)a指向點(diǎn)b的邊的權(quán)重為2.0，而點(diǎn)b指向點(diǎn)a的邊的全中為1/2.0=0.5。
因此我們可以將輸入的表達(dá)式轉(zhuǎn)化為一個(gè)加權(quán)有向圖，而題目的問題則被轉(zhuǎn)化為求兩個(gè)點(diǎn)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(quán)重的乘積。
代碼如下：
public double[] calcEquation(List> equations, double[] values, List> queries) {
//圖的鏈?zhǔn)奖硎痉? Map> pairs = new HashMap<>();
//圖上每條邊的權(quán)重
Map> valuedPairs = new HashMap<>();
for(int i = 0 ; i < equations.size() ; i++) {
//獲取第i個(gè)方程式
List equation = equations.get(i);
String multiplied = equation.get(0);//被除數(shù)
String multiplier = equation.get(1);//除數(shù)
//如果被除數(shù)從來沒有添加到圖中，則將其作為頂點(diǎn)在圖中初始化
if(!pairs.containsKey(multiplied)) {
pairs.put(multiplied, new ArrayList<>());
valuedPairs.put(multiplied, new ArrayList<>());
}
//如果除數(shù)從來沒有添加到圖中，則將其作為頂點(diǎn)在圖中初始化
if(!pairs.containsKey(multiplier)) {
pairs.put(multiplier, new ArrayList<>());
valuedPairs.put(multiplier, new ArrayList<>());
}
//添加邊和邊的權(quán)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(jié)果集
double[] result = new double[queries.size()];
for(int i = 0 ; i(), 1.0);
result[i] = result[i]==0.0 ? -1.0 : result[i];
}
return result;
}
public double dfs(String multiplied, String multiplier, Map> pairs, Map> valuedPairs, Set visited, double curResult) {
//如果圖中不包含該被除數(shù)頂點(diǎn)，則無法獲知該表達(dá)式的值
if(!pairs.containsKey(multiplied)) return 0.0;
//如果再次訪問過該被除數(shù)，說明找到了一條環(huán)路，則該深度優(yōu)先遍歷結(jié)果失敗，直接拋棄
if(visited.contains(multiplied)) return 0.0;
//如果被除數(shù)等于除數(shù)，則返回1.0
if(multiplied.equals(multiplier)) return curResult;
visited.add(multiplied);
//獲得當(dāng)前被除數(shù)的所有鄰接頂點(diǎn)
List multipliers = pairs.get(multiplied);
//獲得所有鄰接邊的權(quán)重
List multiplierValues = valuedPairs.get(multiplied);
double tmp = 0.0;
for(int i = 0 ; i
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