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.

解题思路一:

刚看到题目的时候感觉无从下手,后来仔细理解题意,需要找到first missing positive integer,这个数肯定是1---nums.length+1的一个值,因此我们可以开一个布尔数组,存储i是否存在,JAVA实现如下:

public int firstMissingPositive(int[] nums) {
boolean[] num=new boolean[nums.length];
for (int i = 0; i < nums.length; i++)
if (nums[i] > 0&&nums[i] <= nums.length)
num[nums[i]-1] = true;
for (int i = 0; i < num.length; i++)
if (!num[i])
return i+1;
return nums.length+1;
}

解题思路二:使用bool数组会有O(N)的空间复杂度,直接采用交换的方法可以避免,JAVA实现如下:

static public int firstMissingPositive(int[] nums)  {
if(nums.length==0)
return 1;
for (int i = 0; i < nums.length; ) {
if (nums[i] >= 0 &&nums[i] < nums.length &&nums[i] != i && nums[i] != nums[nums[i]]){
int temp=nums[i];
nums[i]=nums[nums[i]];
nums[temp]=temp;
}
else i++;
}
for (int i = 1; i < nums.length; ++i)
if (nums[i] != i)
return i;
return nums[0] == nums.length ? nums.length + 1 : nums.length;
}

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