LeetCode – First Missing Positive (Java)
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.
Analysis
This problem can solve by using a bucketsort like algorithm. Let's consider finding first missing positive and 0 first. The key fact is that the ith element should be i, so we have:
i==A[i]
A[i]==A[A[i]]
For example, given an array {1,2,0,4}, the algorithm does the following:
int firstMissingPositiveAnd0(int A[]) { int n = A.length; for (int i = 0; i < n; i++) { // when the ith element is not i while (A[i] != i) { // no need to swap when ith element is out of range [0,n] if (A[i] < 0  A[i] >= n) break; //handle duplicate elements if(A[i]==A[A[i]]) break; // swap elements int temp = A[i]; A[i] = A[temp]; A[temp] = temp; } } for (int i = 0; i < n; i++) { if (A[i] != i) return i; } return n; } 
Java Solution
This problem only considers positive numbers, so we need to shift 1 offset. The ith element is i+1.
public int firstMissingPositive(int[] A) { int n = A.length; for (int i = 0; i < n; i++) { while (A[i] != i + 1) { if (A[i] <= 0  A[i] >= n) break; if(A[i]==A[A[i]1]) break; int temp = A[i]; A[i] = A[temp  1]; A[temp  1] = temp; } } for (int i = 0; i < n; i++){ if (A[i] != i + 1){ return i + 1; } } return n + 1; } 
<pre><code> String foo = "bar"; </code></pre>

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