Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
For example,
Given [3,2,1,5,6,4] and k = 2, return 5.
思路:
根据快速排序的partition,每次找到当前选取元素(第一个)在数组中的位置,如果这个位置等于nums.length - 1,就找到了要求的元素;如果小于,那么继续在右边找;如果大于,在左边继续找。
public int findKthLargest(int[] nums, int k) {
if (k < 1 || nums == null) {
return 0;
}
return getKth(nums.length - k +1, nums, 0, nums.length - 1);
}
public int getKth(int k, int[] nums, int start, int end) {
int pivot = nums[end];
int left = start;
int right = end;
while (true) {
while (nums[left] < pivot && left < right) {
left++;
}
while (nums[right] >= pivot && right > left) {
right--;
}
if (left == right) {
break;
}
swap(nums, left, right);
}
swap(nums, left, end);
if (k == left + 1) {
return pivot;
} else if (k < left + 1) {
return getKth(k, nums, start, left - 1);
} else {
return getKth(k, nums, left + 1, end);
}
}
public void swap(int[] nums, int n1, int n2) {
int tmp = nums[n1];
nums[n1] = nums[n2];
nums[n2] = tmp;
}
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