First Missing Positive
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题目:
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
.
思想:
本打算采用常规方法,排序后再进行搜索,发现边界比较复杂,难以处理。后在LeetCode上看到一个答案,具体思想是,先对数组进行重新排序,排序的方法是,如果当前的nums[i]=5,我们就将nums[4]=5;这样一来就可以通过判断nums[i]是否等于i+1来判断是否连续了。
代码:
class Solution{public:int firstMissingPositive(vector<int>& nums) {int len = nums.size();for (int i = 0; i < len; i++) {while (nums[i]>0 && nums[i] <= len && nums[nums[i] - 1] != nums[i])swap(nums[i], nums[nums[i] - 1]);}for (int i = 0; i < len; i++){if (nums[i] != i + 1)return i + 1;}return len + 1;}};
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