|
37 | 37 |
|
38 | 38 | // @lc code=start |
39 | 39 | import ( |
40 | | - "container/heap" |
| 40 | + // "container/heap" |
| 41 | + "math/rand" |
41 | 42 | ) |
42 | 43 |
|
43 | | -type myHeap []int |
| 44 | +// type myHeap []int |
44 | 45 |
|
45 | | -func (h *myHeap) Less(i, j int) bool { |
46 | | - return (*h)[i] < (*h)[j] |
47 | | -} |
| 46 | +// func (h *myHeap) Less(i, j int) bool { |
| 47 | +// return (*h)[i] < (*h)[j] |
| 48 | +// } |
48 | 49 |
|
49 | | -func (h *myHeap) Swap(i, j int) { |
50 | | - (*h)[i], (*h)[j] = (*h)[j], (*h)[i] |
51 | | -} |
| 50 | +// func (h *myHeap) Swap(i, j int) { |
| 51 | +// (*h)[i], (*h)[j] = (*h)[j], (*h)[i] |
| 52 | +// } |
| 53 | + |
| 54 | +// func (h *myHeap) Len() int { |
| 55 | +// return len(*h) |
| 56 | +// } |
52 | 57 |
|
53 | | -func (h *myHeap) Len() int { |
54 | | - return len(*h) |
| 58 | +// func (h *myHeap) Pop() (v interface{}) { |
| 59 | +// *h, v = (*h)[:h.Len()-1], (*h)[h.Len()-1] |
| 60 | +// return |
| 61 | +// } |
| 62 | + |
| 63 | +// func (h *myHeap) Push(v interface{}) { |
| 64 | +// *h = append(*h, v.(int)) |
| 65 | +// } |
| 66 | + |
| 67 | +func findKthLargest(nums []int, k int) int { |
| 68 | + // h := make(myHeap, 0) |
| 69 | + // heap.Init(&h) |
| 70 | + // for _, cur := range nums { |
| 71 | + // if h.Len() >= k { |
| 72 | + // if cur <= h[0] { |
| 73 | + // continue |
| 74 | + // } |
| 75 | + // heap.Pop(&h) |
| 76 | + // } |
| 77 | + // heap.Push(&h, cur) |
| 78 | + // } |
| 79 | + // return h[0] |
| 80 | + s := &solution{nums: nums} |
| 81 | + return s.call(len(nums)-k, 0, len(nums)-1) |
55 | 82 | } |
56 | 83 |
|
57 | | -func (h *myHeap) Pop() (v interface{}) { |
58 | | - *h, v = (*h)[:h.Len()-1], (*h)[h.Len()-1] |
59 | | - return |
| 84 | +type solution struct { |
| 85 | + nums []int |
60 | 86 | } |
61 | 87 |
|
62 | | -func (h *myHeap) Push(v interface{}) { |
63 | | - *h = append(*h, v.(int)) |
| 88 | +// 用快排的分组思想,如果分组的界限是k,就可以返回了,否则继续2分。 |
| 89 | +func (s *solution) call(k, left, right int) int { |
| 90 | + idx := s.partition(left, right) |
| 91 | + if k == idx { |
| 92 | + return s.nums[k] |
| 93 | + } |
| 94 | + if k > idx { |
| 95 | + return s.call(k, idx+1, right) |
| 96 | + } |
| 97 | + return s.call(k, left, idx-1) |
64 | 98 | } |
65 | 99 |
|
66 | | -func findKthLargest(nums []int, k int) int { |
67 | | - h := make(myHeap, 0) |
68 | | - heap.Init(&h) |
69 | | - for _, cur := range nums { |
70 | | - if h.Len() >= k { |
71 | | - if cur <= h[0] { |
72 | | - continue |
73 | | - } |
74 | | - heap.Pop(&h) |
| 100 | +// 左闭右闭 |
| 101 | +func (s *solution) partition(i, j int) int { |
| 102 | + if i >= j { |
| 103 | + return i |
| 104 | + } |
| 105 | + idx := rand.Intn(j-i) + i |
| 106 | + s.nums[idx], s.nums[i] = s.nums[i], s.nums[idx] |
| 107 | + d := s.nums[i] |
| 108 | + for i < j { |
| 109 | + for i < j && s.nums[j] >= d { |
| 110 | + j-- |
| 111 | + } |
| 112 | + s.nums[i] = s.nums[j] |
| 113 | + for i < j && s.nums[i] < d { |
| 114 | + i++ |
75 | 115 | } |
76 | | - heap.Push(&h, cur) |
| 116 | + s.nums[j] =s.nums[i] |
77 | 117 | } |
78 | | - return h[0] |
| 118 | + s.nums[i] = d |
| 119 | + return i |
79 | 120 | } |
80 | 121 |
|
81 | 122 | // @lc code=end |
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