每日算法 230311

题目

面试题 17.05. 字母与数字

难度中等153

给定一个放有字母和数字的数组，找到最长的子数组，且包含的字母和数字的个数相同。

返回该子数组，若存在多个最长子数组，返回左端点下标值最小的子数组。若不存在这样的数组，返回一个空数组。

示例 1:

输入: ["A","1","B","C","D","2","3","4","E","5","F","G","6","7","H","I","J","K","L","M"]

输出: ["A","1","B","C","D","2","3","4","E","5","F","G","6","7"]

示例 2:

输入: ["A","A"]

输出: []

提示：

• array.length <= 100000

来源：LeetCode（力扣）

思路

遍历整个数组得到字母的个数和数字的个数，最长的肯定是和两倍的数字和字母之中最小个数相等，如果其中有一个为0则返回空数组

解题

class Solution {

    public static boolean isInteger(String str) {
        Pattern pattern = Pattern.compile("^[-\\+]?[\\d]*$");
        return pattern.matcher(str).matches();
    }

    public String[] findLongestSubarray(String[] array) {

        int num = 0;
        int abc = 0;

        //遍历数组，获取字母和数字的个数
        for (int i = 0; i < array.length; i++) {
            if (isInteger(array[i])) {
                num ++;
            } else {
                abc ++;
            }
        }

        System.out.println(num);
        System.out.println(abc);

        //存在为0的则返回空数组
        if(num ==0 || abc == 0){
            String[] strings = new String[0];
            return strings;
        }

        //返回的数组长度
        int l = Math.min(abc,num) * 2;

        String[] s = new String[l];

        //滑动窗口
        //内部相同即可
        for (int i = 0; i < array.length - l + 1; i++) {

            int num1 = 0;
            int abc1 = 0;

            for (int j = 0; j < l; j++) {

                s[j] = array[i + j];

                if (isInteger(s[j])) {
                    num1 ++;
                }
            }

            if (num1 == l /2){
                break;
            }

        }

        return s;

    }
}

想错了。。。

看了解答了解了一点前缀和知识和判断字母数字的方法

class Solution {
    public String[] findLongestSubarray(String[] array) {
        Map<Integer, Integer> indices = new HashMap<Integer, Integer>();
        indices.put(0, -1);
        int sum = 0;
        int maxLength = 0;
        int startIndex = -1;
        int n = array.length;
        for (int i = 0; i < n; i++) {
            if (Character.isLetter(array[i].charAt(0))) {
                sum++;
            } else {
                sum--;
            }
            if (indices.containsKey(sum)) {
                int firstIndex = indices.get(sum);
                if (i - firstIndex > maxLength) {
                    maxLength = i - firstIndex;
                    startIndex = firstIndex + 1;
                }
            } else {
                indices.put(sum, i);
            }
        }
        if (maxLength == 0) {
            return new String[0];
        }
        String[] ans = new String[maxLength];
        System.arraycopy(array, startIndex, ans, 0, maxLength);
        return ans;
    }
}


//判断是否是字母可以用>="A"

总结

还是太菜，多多学习