LeetCode 41. First Missing Positive-优快云博客

本文链接：https://blog.youkuaiyun.com/gq__97/article/details/92772191

博客围绕寻找未排序整数数组中缺失的最小正整数展开。要求算法时间复杂度为O(n)且使用常量额外空间。给出了三种解决思路，一是排序后二分查找再顺序查找；二是用HashMap桶存储数字后查找；三是学习大佬的交换排序法，时间复杂度为O(n)。

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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

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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;
    }
}