# O(n) time without extra space Java Solution

• ``````/**
* @author shanhq96@gmail.com & qf0302yty@126.com & 498357435@qq.com
* @ClassName 41_First Missing Positive
* @Description 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.
* @date 2017/3/28-14:10
*/
public class Solution {
public int firstMissingPositive(int[] nums) {
int result = 1;
for (int i = 0; i < nums.length; i++) {
if (nums[i] < 0 || nums[i] > nums.length ) {//由于数组长度为n,故第一个未出现的正整数一定在1~n+1中.故可将范围外的数置为0
nums[i] = 0;
}
}
for (int i = 0; i < nums.length; i++) {
if ((nums[i] % (nums.length + 1)) != 0) {//将出现在数组内的1~n的值进行标记.标记方法为+n+1
nums[nums[i] % (nums.length + 1) - 1] += (nums.length + 1);
}
}
int j = 0;
for (; j< nums.length; j++) {
nums[j] = nums[j] / (nums.length + 1);//根据标记结果计算数组中1~n的数出现的次数,
if (nums[j] == 0){//数组中第一个为0的元素的下标加1即为结果.
break;
}
}
result = j+1;
return result;
}
}
``````

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