描述 Given an unsorted integer array, find the first missing positive integer.
For example, Given [1,2,0] return 3, and [3,4,-1,1] return 2.
Your algorithm should run in O(n) time and uses constant space.
分析 本質(zhì)上是桶排序 (bucket sort)，每當 nums[i]!= i+1 的時候，將 nums[i] 與 nums[nums[i]-1] 交換，直到無法 交換為止，終止條件是 nums[i]== nums[nums[i]-1]。
代碼
class Solution {public: int firstMissingPositive(vector<int>& nums) { bucket_sort(nums); const int n = nums.size(); for (size_t i = 0; i < n; ++i) { if (nums[i] != i + 1) return i + 1; } return n + 1; }PRivate: static void bucket_sort(vector<int>& nums) { const int n = nums.size(); for (size_t i = 0; i < n; ++i) { while (nums[i] != i + 1) { if (nums[i] <= 0 || nums[i] > n || nums[i] == nums[nums[i] - 1]) break; swap(nums[i], nums[nums[i] - 1]); } } }};新聞熱點
疑難解答
