给一整数无序数组，找到第一个缺失的正整数，时间复杂度为n

//    5.7
//    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 int firstMissingPositive(int[] nums) {
        int idx = 0;
        while (idx < nums.length) {
            if (nums[idx] == idx + 1 || nums[idx] <= 0 || nums[idx] > nums.length) {
                idx++;
            } else {
                if (nums[idx] == nums[nums[idx] - 1]) {
                    idx++;
                } else {
                    swip(nums, idx, nums[idx] - 1);
                }
            }
        }
        idx = 0;
        while (idx < nums.length && nums[idx] == idx + 1) {
            idx++;
        }
        return idx + 1;
    }
    void swip(int[] ary,int x,int y) {
        int temp = ary[x];
        ary[x] = ary[y];
        ary[y] = temp;
    }