【剑指offer】之最小的k个数

//求出最小的k个值（时间复杂度o(n)）
private static void getLeastNumbers(int []nums, int k) {

int []obs = new int[nums.length]; //这样就不会改变nums里面的值的位置。
for(int i=0;i<nums.length;i++) {
obs[i] = nums[i];
}

if(obs.length < 0 || k<=0)
return ;

int start = 0;
int end = obs.length - 1;
int index = Partition(obs,start,end);
while(index != k-1) {
if(index > k-1) {
end = index -1;
index = Partition(obs,start,end);
}
else {
start = index + 1;
index = Partition(obs,start,end);
}
}

//输出索要的k个值
for(int i=0;i<k;i++) {
System.out.print(obs[i]+",");
}
}

private static int Partition(int []obs, int start ,int end) {

int j = 0;
if(start < end) {
int i = start ;
j = end + 1;
//			int temp = obs[start];

while (true) {
while (obs[start] > obs[++i] && i < end)
;
while (obs[start] < obs[--j] && j > start)
;
if (i < j) {
swap(obs, i, j);
} else {
break;
}
}

swap(obs,start,j);
}

return j;
}

private static void swap(int[] obs, int i, int j) {
int swap = obs[i];
obs[i] = obs[j];
obs[j] = swap;
}