LeetCode 41. First Missing Positive

Description

Given an unsorted integer array, find the smallest missing positive integer.

Note

Your algorithm should run in O(n) time and uses constant extra space.

Example

Code

• java

class Solution {
    public int getFirstPositivePosition(int[] nums) {
        int left = 0, right = nums.length - 1;
        while(left <= right) {
            int mid = (left + right) / 2;
            if(nums[mid] <= 0) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
    public int firstMissingPositive2(int[] nums) {
        Arrays.sort(nums);
        int pos = getFirstPositivePosition(nums);
        int result = 1;
        for(int i = pos; i < nums.length; i++) {
            if(nums[i] != result) {
                break;
            }
            while(i + 1 < nums.length && nums[i+1] == nums[i]) {
                i++;
            }
            result++;
        }
        return result;
    }
    public int firstMissingPositive3(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        for(Integer i : nums) {
            if(i <= 0) continue;
            map.put(i, 1);
        }
        int result = 1;
        while(result <= map.size()) {
            if(map.get(result) == null) return result;
            result++;
        }
        return result;
    }
    public int firstMissingPositive(int[] nums) {
        int len = nums.length;
        for(int i = 0; i < len; i++) {
            while(nums[i] > 0 && nums[i] <= len && nums[nums[i] - 1] != nums[i]) {
                swap(nums, nums[i] -1, i);
            }
        }
        for(int i = 0; i < len; i++) {
            if(nums[i] != i + 1) return i + 1;
        }
        return len + 1;
    }
    public void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

Conclusion

• 第一个想法是排序O(nlogn)后，二分O(nlogn)找到第一个正数，然后顺序查找O(n)第一个Missing的数，结果发现里面还有重复的数字，需要处理一下
• 第二个HashMap桶存储出现的数字，然后从1开始查找
• 第三次学习大佬交换排序的方法，下标范围0-len-1，要交换的数字范围1-len，时间复杂度O(n)