如何按照子数组长度对二维数组进行降序排序

完全可以实现该需求，排序逻辑只需要将子数组的长度作为排序权重，指定降序规则即可，排序过程不会修改子数组内部的元素顺序，完全匹配你给出的示例效果。

示例说明 #

输入的二维数组：

[["1","b","g"],["e","3"],["r","2","9","a"],["2"]]

各子数组的长度分别为3、2、4、1，按长度从大到小排序后，输出结果为：

[["r","2","9","a"],["1","b","g"],["e","3"],["2"]]

常用语言实现示例 #

• Python 实现：

original_arr = [["1","b","g"],["e","3"],["r","2","9","a"],["2"]]
# key指定排序依据为子数组长度，reverse=True表示降序
sorted_arr = sorted(original_arr, key=lambda item: len(item), reverse=True)

• JavaScript 实现：

const originalArr = [["1","b","g"],["e","3"],["r","2","9","a"],["2"]];
// 排序规则：长度大的子数组排在前面
const sortedArr = originalArr.sort((a, b) => b.length - a.length);

• Java 实现（JDK8+）：

import java.util.Arrays;
import java.util.Comparator;
import java.util.List;

public class SortTest {
    public static void main(String[] args) {
        List<List<String>> originalArr = Arrays.asList(
                Arrays.asList("1","b","g"),
                Arrays.asList("e","3"),
                Arrays.asList("r","2","9","a"),
                Arrays.asList("2")
        );
        // 按子数组长度降序排序
        originalArr.sort(Comparator.comparingInt(List::size).reversed());
    }
}

内容的提问来源于stack exchange，提问作者Zhu