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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.
对于元素为n的数组A, 正数个数 <= n, 那么第一个miss的正数必然 <= n+1。这样，可以把负数置成n+2，不影响结果。
假设数字k 属于 [1, n]，且在数组A中的j位置出现，即A[j] = k。把数组A中所有的k置成-k, 那么留下第一个正数的位置j
即为第一个miss的正数。
1 int firstMissingPositive(int A[], int n) 2 { 3 int i = 0, temp = 0, inf = n + 2; 4 5 for (i = 0; i < n; i++) 6 A[i] = (A[i] <= 0) ? inf : A[i]; 7 8 for (i = 0; i < n; i++) 9 { 10 temp = A[i] > 0 ? A[i] : -A[i]; 11 if (temp > n) 12 continue; 13 if (A[temp - 1] > 0) /* in case that A has duplicate value and A[temp - 1] will * -1 twice. */ 14 A[temp - 1] *= -1; 15 } 16 17 for (i = 0; i < n; i++) 18 { 19 if (A[i] > 0) 20 return i + 1; 21 } 22 23 return i + 1; 24 }
leetcode .First Missing Positive
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原文地址：http://www.cnblogs.com/ym65536/p/4095897.html
