标签：
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.
Analyse:
1. sort and locate the positive first number.
Runtime: 4ms.
1 class Solution { 2 public: 3 int firstMissingPositive(vector<int>& nums) { 4 sort(nums.begin(), nums.end()); 5 6 int positive = 1; 7 for(int i = 0; i < nums.size(); i++){ 8 if(nums[i] > 0){ 9 if(nums[i] != positive) return positive; 10 for(int j = i; j < nums.size() - 1; j++){ 11 if(nums[j + 1] - nums[j] <= 1) continue; 12 return nums[j] + 1; 13 } 14 return nums[nums.size() - 1] + 1; 15 } 16 } 17 return positive; 18 } 19 };
2. put nums[i] at (i + 1)-th index. For each element satisfying this requirement, we put it in the right position. For those not, if they are large than the vector‘s size or are negative numbers or equals to nums[nums[i] - 1], we can ignore this element and deal with the next element.
Runtime: 4ms.
1 class Solution { 2 public: 3 int firstMissingPositive(vector<int>& nums) { 4 int n = nums.size(); 5 6 for(int i = 0; i < n; i++){ 7 while(nums[i] != i + 1){ 8 if(nums[i] < 0 || nums[i] > n || nums[i] == nums[nums[i] - 1]) break; 9 swap(nums[i], nums[nums[i] - 1]); 10 } 11 } 12 13 for(int j = 0; j < n; j++){ 14 if(nums[j] != j + 1) return j + 1; 15 } 16 return n + 1; 17 } 18 };
标签：
原文地址：http://www.cnblogs.com/amazingzoe/p/4699228.html
