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Update 47.permutations-ii.md
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problems/47.permutations-ii.md

Lines changed: 3 additions & 4 deletions
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@@ -67,9 +67,8 @@ function backtrack(list, nums, tempList, visited) {
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for (let i = 0; i < nums.length; i++) {
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// 和46.permutations的区别是这道题的nums是可以重复的
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// 我们需要过滤这种情况
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if (visited[i]) continue; // 不能用tempList.includes(nums[i])了,因为有重复
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// visited[i - 1] 这个判断容易忽略
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if (i > 0 && nums[i] === nums[i - 1] && visited[i - 1]) continue;
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if (visited[i]) continue; // 同一个数字不能用两次
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if (i > 0 && nums[i] === nums[i - 1] && visited[i - 1]) continue; // 同样值的数字不能用两次
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visited[i] = true;
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tempList.push(nums[i]);
@@ -112,7 +111,7 @@ class Solution:
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res.append(pre_list)
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else:
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for i in range(len(nums)):
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# 如果是同样的数字,则之前一定已经生成了对应可能
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# 同样值的数字只能使用一次
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if i > 0 and nums[i] == nums[i-1]:
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continue
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p_list = pre_list.copy()

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