leetcode First Missing Positive

最新推荐文章于 2016-05-20 10:41:19 发布

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本文介绍了一种在未排序整数数组中寻找第一个缺失正整数的算法，该算法能在O(n)时间内运行并使用常数空间。通过具体示例展示了如何实现这一算法，并对比了两种不同版本的代码实现，帮助理解关键步骤。

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.

swap

class Solution {
 public:
  int firstMissingPositive(int A[], int n) {
    int i, An = 0;
    for (i = 0; i < n; ++i)
      while (A[i] != i)
        if (A[i] >= 0 && A[i] < n && A[A[i]] != A[i])
          swap(A[i], A[A[i]]);  
        else if (A[i] == n) {
          An = n;
          break;
        }
        else
          break;
    for (i = 1; i < n; ++i)
      if (A[i] != i)
        return i;
    if (An)
      return An + 1;
    else
      return n ? n : 1;
  }
};

Here is a TLE version, find the bug:

class Solution {
 public:
  int firstMissingPositive(int A[], int n) {
    for (int i = 1; i <= n; ++i) {
      while (A[i-1] != i && A[i-1] >= 1 && A[i-1] <= n) {
        swap(A[i-1], A[A[i-1]-1]);
      }
    }      
    for (int i = 1; i <= n; ++i)
      if (A[i-1] != i)
        return i;
    return n+1;
  }
};

AC Code:

class Solution {
 public:
  int firstMissingPositive(int A[], int n) {
    for (int i = 1; i <= n; ++i) {
      while (A[i-1] != i && A[i-1] >= 1 && A[i-1] <= n && A[A[i-1]-1] != A[i-1]) {
        swap(A[i-1], A[A[i-1]-1]);
      }
    }      
    for (int i = 1; i <= n; ++i)
      if (A[i-1] != i)
        return i;
    return n+1;
  }
};