41. First Missing Positive
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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.
题目要求找到数组中缺失的最小的正数,并要求时间复杂度为O(n),空间复杂度为O(1)。所以不能排序,不能新建一个数组记录某个数是否出现,只能对原数组做操作,如果nums[i]大于0且小于等于nums数组的大小(即此值减1能作为索引),则将nums[i]与nums[nums[i]-1]交换(如果这两个值一样就不用管了)。这样就能使数组索引为i处的值为i+1(前提是i+1在原数组中存在)。这样再遍历一次数组,如果某处i的值不为i+1,说明缺失这个值,直接返回i+1,如果遍历完都没有返回,则返回n+1。
代码:
class Solution {public: int firstMissingPositive(vector<int>& nums) { int n = nums.size(); for (int i = 0; i < n; i++) while (nums[i] > 0 && nums[i] <= n && nums[nums[i] - 1] != nums[i]) swap(nums[i], nums[nums[i] - 1]); for (int i = 0; i < n; i++) if (nums[i] != i + 1) return i + 1; return n + 1; }};
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