将各个正数放入相应下标处,使之满足nums[nums[i] - 1] = nums[i],则一遍扫描后值在数组长度范围内的都按序排列了,然后二遍扫描遇到第一个打破秩序的位置i,则 i + 1就是第一个缺失的正数。
[ref]
解析清楚的Code Ganker博客: http://blog.csdn.net/linhuanmars/article/details/20884585
简洁的实现: https://leetcode.com/discuss/24013/my-short-c-solution-o-1-space-and-o-n-time
public class Solution { public int firstMissingPositive2(int[] nums) { if (nums == null) return 1; int N = nums.length; for (int i = 0; i < N; i++) { while (nums[i] > 0 && nums[i] <= N && nums[nums[i] - 1] != nums[i]) { int tmp = nums[nums[i] - 1]; nums[nums[i] - 1] = nums[i]; nums[i] = tmp; } } for (int i = 0; i < N; i++) { if (nums[i] != i + 1) return i + 1; } return N + 1; } public int firstMissingPositive1(int[] nums) { if (nums == null) return 1; int N = nums.length; for (int i = 0; i < N; i++) { if (nums[i] > 0 && nums[i] <= N && nums[nums[i] - 1] != nums[i]) { int tmp = nums[nums[i] - 1]; nums[nums[i] - 1] = nums[i]; nums[i] = tmp; i--; } } for (int i = 0; i < N; i++) { if (nums[i] != i + 1) return i + 1; } return N + 1; } }