package com.code; import java.util.Arrays; public class Test04_1 { public static int solution(int[] A) { int size = A.length; if(size==1){ // handle one element array if(A[0]==1){ return 2; }else{ return 1; } } Arrays.sort(A); // sort by JDK if(A[0]>1){ // handle all elements bigger than 1 return 1; } if(A[size-1]<=0){ // handle all elements are negative return 1; } int i=0; for(i=0;i<size-1;i++){ if(A[i]>=0 && (A[i+1]-A[i]>1)){ // handle no consecutive array which start with 1 return A[i]+1; } } if(i==size-1){ // handle consecutive array , handle negative elements too. return A[size-1]+1>0?A[size-1]+1:1; } return 1; } public static void main(String[] args) { int [] a = {1,1,2,3,4,5}; System.out.println(solution(a)); int [] b = {-4,-2,0,5}; System.out.println(solution(b)); int [] c = {2}; System.out.println(solution(c)); int [] d = {-2,0,100,102,200}; System.out.println(solution(d)); int [] e = {2,4,5,6}; System.out.println(solution(e)); } } /** * * 1. MissingInteger Find the minimal positive integer not occurring in a given sequence. * Write a function: * * class Solution { public int solution(int[] A); } * * that, given a non-empty zero-indexed array A of N integers, returns the minimal positive integer (greater than 0) that does not occur in A. * * For example, given: * * A[0] = 1 A[1] = 3 A[2] = 6 A[3] = 4 A[4] = 1 A[5] = 2 the function should return 5. * * Assume that: * * N is an integer within the range [1..100,000]; each element of array A is an integer within the range [?2,147,483,648..2,147,483,647]. Complexity: * * expected worst-case time complexity is O(N); expected worst-case space complexity is O(N), * beyond input storage (not counting the storage required for input arguments). Elements of input arrays can be modified. * */
时间: 2024-10-06 10:29:40