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LeetCode刷题：41. First Missing Positive

原题链接：

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 Note:

Your algorithm should run in O(n) time and uses constant extra space.

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算法设计

package com.bean.algorithmbasic;

/*

 * 41. First Missing Positive  * 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 FirstMissingPositive {

         public static int findFirstMissingPositive1(int[] nums) {         int n = nums.length;                  // 1. mark numbers (num < 0) and (num > n) with a special marker number (n+1)          // (we can ignore those because if all number are > n then we'll simply return 1)         for (int i = 0; i < n; i++) {             if (nums[i] <= 0 || nums[i] > n) {                 nums[i] = n + 1;             }         }         // note: all number in the array are now positive, and on the range 1..n+1                  // 2. mark each cell appearing in the array, by converting the index for that number to negative         for (int i = 0; i < n; i++) {             int num = Math.abs(nums[i]);             if (num > n) {                 continue;             }             num--; // -1 for zero index based array (so the number 1 will be at pos 0)             if (nums[num] > 0) { // prevents double negative operations                 nums[num] = -1 * nums[num];             }         }                  // 3. find the first cell which isn't negative (doesn't appear in the array)         for (int i = 0; i < n; i++) {             if (nums[i] >= 0) {                 return i + 1;             }         }                  // 4. no positive numbers were found, which means the array contains all numbers 1..n         return n + 1;     }

    public static int findFirstMissingPositive2(int[] target) {

        if (target.length == 0 || target == null)             return 1;         // 把元素放入正确的位置，例如1放在target[0]，2放在target[1]...         for (int i = 0; i < target.length; i++) {             while (target[i] != i + 1) {                 if (target[i] >= target.length || target[i] <= 0 || target[i] == target[target[i] - 1])                     break;                 int temp = target[i];                 target[i] = target[temp - 1];                 target[temp - 1] = temp;             }         }

        for (int i = 0; i < target.length; i++) {

            if (target[i] != i + 1)                 return i + 1;         }         return target.length + 1;     }          public static int findFirstMissingPositive3(int[] nums) {         if(nums.length==0){             return 1;         }         int ans=1;         for(int i=0;i<nums.length;i++){             if(nums[i]==i+1){                 continue;             }             while(nums[i]<=nums.length&&nums[i]>0&&nums[i]!=nums[nums[i]-1]){                 int temp=nums[nums[i]-1];                 nums[nums[i]-1]=nums[i];                 nums[i]=temp;             }         }         int k=0;         for(k=0;k<nums.length;k++){             if(nums[k]!=k+1){                 break;             }         }         return k+1;     }

    public static void main(String[] args) {

        // TODO Auto-generated method stub

        int[] demo = new int[] { 3, 4, -1, 1 };

        // int[] demo=new int[] {1,2,0};

        int result = findFirstMissingPositive3(demo);

        System.out.println("result= " + result);

    }

}

程序运行结果：

result= 2

 

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