Question

有谁能告诉我哪里出错了。以下是我的代码。这里我使用左右滑动窗口解决方案作为String的指针。 “dist_count”包含字符串中唯一字符的计数。

public class Solution {
    public String findSubstring(String s) {
        if (s == null || s.length() == 0)
            return "";
        int dist_count = 0;
        int[] hash = new int[256]; // character hash
        boolean[] visited = new boolean[256];
        // record each character in s to hash
        for (char c : s.toCharArray()) {
            hash[c]++;
        }
        for (char c : s.toCharArray()) {
            if (visited[c] == false) {
                dist_count++;
                visited[c] = true;
            }
        }
        System.out.println(dist_count);
        int left = 0, right = 0, count = 0;
        int min_window = Integer.MAX_VALUE;
        int startIndex = -1;
        while (right < s.length()) {
            if (hash[s.charAt(right)] >= 1) {
                count++;
            }
            hash[s.charAt(right)]--;
            right++;

            if (count == dist_count) {
                while (hash[s.charAt(left)] > 1) {
                    if (hash[s.charAt(left)] > 1)
                        hash[s.charAt(left)]--;
                    left++;
                }
                int window = right - left + 1;
                if (min_window > window) {
                    min_window = window;
                    startIndex = left;
                }
            }
        }
        return s.substring(startIndex, startIndex + min_window);
    }

    public static void main(String args[]) {
        Solution ob = new Solution();
        String str = ob.findSubstring("abdacbfcabegbdfcadcefa");
        System.out.println(str);
    }
}

Answer 1

您的算法是正确的，但实施中存在一些问题。您不应该使用原始字符串中的哈希剩余部分。请查看以下代码：

android studio 3.0.1

Working code上述更改

具有原始字符串的所有不同字符的最小长度子字符串的滑动窗口解决方案

1 个答案: