41. First Missing Positive
Given an unsorted integer array, find the smallest missing positive integer.
Example 1:
Input: [1,2,0]
Output: 3
Example 2:
Input: [3,4,-1,1]
Output: 2
Example 3:
Input: [7,8,9,11,12]
Output: 1
我的基本思路是一个长度为l的数组,假设从0开始存放,最多存放l-1个正整数,用布尔型数组去判断是否存在这个正整数,如果都存在,那就返回l
class Solution {
public int firstMissingPositive(int[] nums) {
boolean[] res=new boolean[nums.length+1];
for(int i=0;i<nums.length;i++){
if(nums[i]>0&&nums[i]<res.length){
res[nums[i]]=true;
}
}
for(int i=1;i<res.length;i++){
if(!res[i])
return i;
}
return res.length;
}
}
借鉴其它代码,可以采用交换的方式,降低空间复杂度
class Solution {
public int firstMissingPositive(int[] nums) {
int i = 0;
while (i < nums.length) {
if (nums[i] == i + 1 || nums[i] <= 0 || nums[i] > nums.length) {
i++;
} else if (nums[nums[i] - 1] != nums[i]) {
swap(nums, i, nums[i] - 1);
} else {
i++;
}
}
i = 0;
while (i < nums.length && nums[i] == i + 1)
i++;
return i + 1;
}
public void swap(int[] nums,int i,int j) {
int tmp=nums[i];
nums[i]=nums[j];
nums[j]=tmp;
}
}