我在LeetCode上解决了您需要的问题,但目前我的代码占用了太多内存(~37 my ),与其他提交的代码相比运行得非常慢。我的解决办法是: def lengthOfLongestSubstring(self, s: str) -> int: for a in range(len(s)): strings.append(len(s
我知道使用HashSet来获取唯一计数,但是它占用了太多的内存。beforeUsedMem = Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory(); for (int index = 0; index < 1000000; index++) {
let strings = vec!["Rust", "Rest", "Rust"]; // I want to find "Rust" in this case
.into_iter().find(|x| o.into_iter().filter(|y| x == y).count() >= 2)
// sorry o ^ here is supposed to be s
当我试图从一个可能不止一次包含该元素的OutOfMemoryError中删除特定元素时,我会得到一个ArrayList和其他ArrayList错误。 ArrayList<String> s = new ArrayList<String> (); /* List which will have some of its elements removed */