[LeetCode] [C] 41. First Missing Positive

In this program,

I used Bubble Sort  to sort the array first.

Then, I declared a variable "count" to represent the First Missing Positive Integer.

(Because the way I used it is similar to when I am counting something.)

Simply speaking,

I start counting when nums[i] > 0.

(We are finding the first missing positive integer, so the numbers smaller than 0 make no sense.)

After submitting once,

I found that there could be duplicate numbers.

So if (nums[i]==nums[i-1]) count--;

You'll notice that the value difference between nums[i] and nums[i+1] is always 0 or 1,

if it's not,

there must be a missing positive integer.

As count++ and count-- goes,

when nums[i]!=count,

the result comes out.

For example,

nums: [-1, 0, 1, 2, 2, 4]

First missing positive integer: 3

when nums[i]>0, counting starts, and "i" is still increasing.

i=2: nums[i]=1, count=1;    i++,count++;

i=3: nums[i]=2, count=2;    i++,count++;

i=4: nums[i]=2, count=2;    i++,count++,count--; (at this moment, nums[i]==nums[i-1], so count--)

i=5: nums[i]=4, count=3;

Here, nums[i]!=count, the answer comes.

The Runtime of the program is  3 ms.

The length of the code is  465 Bytes.

int firstMissingPositive(int* nums, int numsSize) {    int i,j,tmp,count=1;    for (i=0; i<numsSize; i++) {        for (j=i+1; j<numsSize; j++) {if (nums[i]>nums[j]) {tmp=nums[i];nums[i]=nums[j];nums[j]=tmp;}        }    }    for (i=0; i<numsSize; i++) {        if (nums[i]>0) {if (i>0)if (nums[i]==nums[i-1]) count--;            if (nums[i]==count) count++;            else return count;        }    }    return count;}