- 给定一个整数数组 a，其中1 ≤ a[i] ≤ n （n为数组长度）,
其中有些元素出现两次而其他元素出现一次,找到所有出现两次的元素
[要求Tc: O(n) Sc:O(1)]
例:
输入:
[4,3,2,7,8,2,3,1]
// [7,3,2,4,8,2,3,1]
// [3,3,2,4,8,2,7,1]
// [2,3,3,4,8,2,7,1]
// [2,3,3,4,1,2,7,8]
// [1,2,3,4,3,2,7,8]
输出:
[2,3]
主要麻烦的点在于空间需要Sc: O(1)
const appearTwice = function(ary) {
const aryLen = ary.length;
for (let i = 0; i < aryLen; i += 1) {
const item = ary[i];
if (ary[item-1] !== ary[i]) {
[ary[item-1], ary[i]] = [ary[i], ary[item-1]];
i --;
}
}
for (let i = 0; i < aryLen; i += 1) {
const item = ary[i];
if (item - 1 !== i) {
ary.push(item);
}
}
return ary.slice(aryLen);
}
const ary = [4,3,2,7,8,2,3,1];
console.log(appearTwice(ary));
