问题：
方法： 使用DFS遍历所有可能经过的节点，然后通过SET去掉重复的Case，最终可得到所有符合要求的结果。
package com.eric.leetcode
class SubsetsII {
fun subsetsWithDup(nums: IntArray): List<List<Int>> {
val result = mutableSetOf<List<Int>>()
result.add(mutableListOf())
for (index in nums.indices) {
dfs(nums, index, mutableListOf(), result)
}
return result.toList()
}
private fun dfs(nums: IntArray, cur: Int, list: MutableList<Int>, result: MutableSet<List<Int>>) {
val newList = mutableListOf<Int>()
newList.addAll(list)
newList.add(nums[cur])
newList.sort()
result.add(newList)
for (index in cur + 1..nums.lastIndex) {
dfs(nums, index, newList, result)
}
}
}
fun main() {
val input = intArrayOf(1, 2, 2)
val subsetsII = SubsetsII()
subsetsII.subsetsWithDup(input)
}
有问题随时沟通
