我試圖在數(shù)組中找到子數(shù)組。它僅適用于一個(gè)子數(shù)組，但我希望如果有多個(gè)子數(shù)組，它會(huì)返回最后一個(gè)的索引。例如，對(duì)于 [3,4,1,2,0,1,2,5,6] 和 [1,2] 應(yīng)該返回 5。public int FindArray(int[] array, int[] subArray) { //throw new NotImplementedException(); int y=0; int index=0; bool find= false; for(int x=0;x< array.Length && y< subArray.Length;) { if(array[x]!= subArray[y]) { if(find==true) { y=0; index=x; } else { x++; y=0; index=x; } } else { find=true; x++; y++; } } if(y==subArray.Length) return index; else return -1; }}
1 回答
一只甜甜圈
TA貢獻(xiàn)1836條經(jīng)驗(yàn) 獲得超5個(gè)贊
public int FindLast(int[] haystack, int[] needle)
{
// iterate backwards, stop if the rest of the array is shorter than needle (i >= needle.Length)
for (var i = haystack.Length - 1; i >= needle.Length - 1; i--)
{
var found = true;
// also iterate backwards through needle, stop if elements do not match (!found)
for (var j = needle.Length - 1; j >= 0 && found; j--)
{
// compare needle's element with corresponding element of haystack
found = haystack[i - (needle.Length - 1 - j)] == needle[j];
}
if (found)
// result was found, i is now the index of the last found element, so subtract needle's length - 1
return i - (needle.Length - 1);
}
// not found, return -1
return -1;
}
作為一個(gè)可運(yùn)行的小提琴：https ://dotnetfiddle.net/TfjPuY
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