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First Missing Positive - Leetcode
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/* | |
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. | |
*/ | |
public class Solution { | |
public int firstMissingPositive(int[] A) { | |
// Start typing your Java solution below | |
// DO NOT write main() function | |
// ========= Time complexity: O(n) =========// | |
int i = 0; | |
while(i < A.length) { | |
if(A[i] != i+1) { | |
if(0 <= A[i]-1 && A[i]-1 < A.length) { | |
int tmp = A[A[i]-1]; | |
if(tmp != A[i]) { | |
A[A[i]-1] = A[i]; | |
A[i] = tmp; | |
} else { | |
// deal with this case: [2,2] | |
i++; | |
} | |
} else { | |
// deal with this case: [0,1] or [3,1] | |
i++; | |
} | |
} else { | |
// deal with case when A[i] == i+1, like [1,2] | |
i++; | |
} | |
} | |
int j = 0; | |
for(; j < A.length; j++) { | |
if(A[j] != j+1) return j+1; | |
} | |
return j+1; | |
// ========= Time complexity: O(n*log(n)) =======// | |
if(A.length == 0) return 1; | |
Arrays.sort(A); // sort the array | |
int idx = 0; | |
int offset = 0; | |
while(offset < A.length && A[offset] <= 0) { | |
// check the offset to find the starting point | |
offset++; | |
} | |
int i = 1; | |
for(; offset < A.length; offset++) { | |
if(offset > 0 && A[offset] == A[offset-1]) continue; | |
if(A[offset] != i) break; // if it does not follow the rule when it is not increased by 1. | |
i++; | |
} | |
return i; | |
} | |
} |
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