leetcode-java/src/FirstMissingPositive41.java at master · fluency03/leetcode-java · GitHub

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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.
 */


importjava.util.ArrayList;
importjava.util.List;


publicclassFirstMissingPositive41 {
publicintfirstMissingPositive(int[] nums) {

List<Boolean> isMissing = newArrayList<>();
isMissing.add(false);

intfirstMissing = 1;

for (inti = 0; i < nums.length; i++) {
intnow = nums[i];
if (now > 0) {
try {

if (isMissing.get(now)) {
isMissing.set(now, false);
 }

 } catch (IndexOutOfBoundsExceptione) {


while (isMissing.size() < now) {
isMissing.add(true);
 }

isMissing.add(false);
 }


 }
try {
while (!isMissing.get(firstMissing)) {
firstMissing++;
 }
 } catch (IndexOutOfBoundsExceptione) {
isMissing.add(true);
 }


 }

returnfirstMissing;

 }


/**
 *
 */
publicintfirstMissingPositive2(int[] nums) {
if (nums == null || nums.length == 0) {
return1;
 }

for (inti = 0; i < nums.length; ++i) {
while (nums[i] > 0 && nums[i] <= nums.length && nums[i] - 1 != i) {
inttmp = nums[nums[i] - 1];
if (tmp == nums[i]) {
break;
 }
nums[nums[i] - 1] = nums[i];
nums[i] = tmp;
 }
 }

for (inti = 0; i < nums.length; ++i) {
if (nums[i] != i + 1) {
returni + 1;
 }
 }

returnnums.length + 1;
 }


/**
 * https://leetcode.com/problems/first-missing-positive/discuss/17083/O(1)-space-Java-Solution
 */
publicintfirstMissingPositive3(int[] A) {
inti = 0;
while(i < A.length){
if(A[i] == i+1 || A[i] <= 0 || A[i] > A.length) i++;
elseif (A[A[i]-1] != A[i]) swap(A, i, A[i]-1);
elsei++;
 }
i = 0;
while(i < A.length && A[i] == i+1) i++;
returni+1;
 }

privatevoidswap(int[] A, inti, intj){
inttemp = A[i];
A[i] = A[j];
A[j] = temp;
 }


publicintfirstMissingPositive4(int[] nums) {
if (nums == null || nums.length == 0) return1;
intN = nums.length;
for (inti=0; i<N; i++) {
intidx = i;
intval = nums[idx];
while (idx >= 0 && idx < N && val > 0 && val <= N && nums[idx] != idx + 1) {
intnewVal = nums[val - 1];
nums[val - 1] = val;
idx = newVal - 1;
val = newVal;
 }
 }

for (inti=0; i<N; i++) {
if (nums[i] != i + 1) {
returni + 1;
 }
 }
returnN + 1;
 }

}