//时间复杂度O(NlogN),空间复杂度O(1)
void solution2(int* src, int len) {
//先快排,使所有相同项聚集
qsort(src, len, sizeof(int), cmp);
int prev = 0;
//若带放入元素与上一个放入元素相同则跳过(重复项)
for (int i = 1; i < len; ++i) {
if (src[prev] != src[i]) {
src[prev + 1] = src[i];
prev++;
}
}
for (int i = 0; i < prev + 1; ++i) {
cout << src[i] << ' ';
}
}