leetcode-剑指53-II-OK

可以用二分优化
address

int missingNumber(int* nums, int numsSize){
	int guiwei(int shu){
		int a= nums[shu];
		nums[shu] = shu;
		if((a>=0)&&(a<numsSize))
			guiwei(a);
		return a;
	}
	for(int i = 0; i< numsSize; i++){
        if(nums[i] != i){
            if((nums[i]>=0)&&(nums[i]<numsSize))
            	guiwei(nums[i]);
        }
    }
    for(int i = 0; i< numsSize; i++){
        if(nums[i] != i)
        	return i;
    }
    return numsSize;
}