6037. 按奇偶性交换后的最大数字

执行用时：2 ms, 在所有 Java 提交中击败了100.00% 的用户
内存消耗：38.6 MB, 在所有 Java 提交中击败了100.00% 的用户

题目链接：https://leetcode-cn.com/problems/largest-number-after-digit-swaps-by-parity/

解题思路

将下标存入list，同时收集整数，排序后依次放回

代码

class Solution {
    public int largestInteger(int num) {
        char[] arr = String.valueOf(num).toCharArray();
        int len = arr.length;
        int[] res = new int[len];
        int ans = 0;
        ArrayList<Integer> oddIdx = new ArrayList<>();
        ArrayList<Integer> odd = new ArrayList<>();
        ArrayList<Integer> evenIdx = new ArrayList<>();
        ArrayList<Integer> even = new ArrayList<>();
        for (int i = 0; i < len; i++) {
            int cur = arr[i] - '0';
            if (cur % 2 == 1) {
                odd.add(cur);
                oddIdx.add(i);
            } else {
                even.add(cur);
                evenIdx.add(i);
            }
        }
        odd.sort(Comparator.reverseOrder());
        even.sort(Comparator.reverseOrder());
        for (int i = 0; i < oddIdx.size(); i++) {
            res[oddIdx.get(i)] = odd.get(i);
        }
        for (int i = 0; i < evenIdx.size(); i++) {
            res[evenIdx.get(i)] = even.get(i);
        }
        for (int i = 0; i < res.length; i++) {
            ans = ans * 10 + res[i];
        }
        return ans;
    }
}