工作答案:http://jsfiddle.net/y3h9L/
行,所以如果我明白量要求的是,被假定爲具有偶數x中的元素,y對的一維陣列。
A = [1,2, 1,3, 2,3, 9,10, 9,11, 10,11]
// output should be
[ [1,2,3], [9,10,11] ]
// but if you add an extra pair that links the two halves, say add 2,11
A2 = [1,2, 1,3, 2,3, 9,10, 9,11, 10,11, 2,11]
// then all are related so output should be
[ [1,2,3,9,10,11] ]
我毫不漂亮的向上或優化下面的代碼,但它的工作原理:
// single dimensional array of x,y pairs
var A = [1,2, 1,3, 2,3, 9,10, 9,11, 10,11];
// create a working copy of A so that we can remove elements
// and still keep the original A intact.
var workingCopy = A.slice(0, A.length),
matchedPairs = [],
currentMatches,
finalCombinations = [],
x, y, i, j,
tempArray;
while (workingCopy.length > 0) {
currentMatches = [];
currentMatches.push([workingCopy.shift(),workingCopy.shift()]);
workingCopyLoop:
for (x=0,y=1; x < workingCopy.length;) {
for (i=0; i < currentMatches.length; i++){
if (workingCopy[x] === currentMatches[i][0]
|| workingCopy[y] === currentMatches[i][1]) {
currentMatches.push([workingCopy.shift(),workingCopy.shift()]);
// go back to the beginning of workingCopyLoop
x=0;
y=1;
continue workingCopyLoop;
}
}
x += 2;
y += 2;
}
matchedPairs.push(currentMatches);
}
for (i=0; i<matchedPairs.length; i++){
tempArray = [];
for (j=0; j<matchedPairs[i].length; j++) {
// I assume you have a new enough version of JS that you have Array.indexOf()
if (-1 === tempArray.indexOf(matchedPairs[i][j][0]))
tempArray.push(matchedPairs[i][j][0]);
if (-1 === tempArray.indexOf(matchedPairs[i][j][1]))
tempArray.push(matchedPairs[i][j][1]);
}
finalCombinations.push(tempArray);
}
for (i=0; i<finalCombinations.length; i++)
console.log(finalCombinations[i]);
// console.log shows that finalCombinations = [ [1,2,3], [9,10,11] ]
如果不是很明顯這是如何工作,用調試器遵循它通過和/或鉛筆和紙。
這是JavaScript的嗎?你的意思是到處使用括號[],而不是像你一樣使用括號()? – BobS
你是否有充分的理由認爲你將永遠 - 或通常 - 能夠以這種方式打破這些事情?如果你有[[1,2],[1,3],[9,3],[9,10]]會怎麼樣?在這種情況下你期望的結果是什麼? – BobS
JavaScript,是的! – JsusSalv