|
62 | 62 |
|
63 | 63 | <!-- 这里可写通用的实现逻辑 -->
|
64 | 64 |
|
| 65 | +**方法一:优先队列(大小根堆)+ 前后缀和 + 枚举分割点** |
| 66 | + |
| 67 | +题目实际上等价于在 $nums$ 中找到一个分割点,将数组分成左右两部分,在前一部分中选取最小的 $n$ 个元素,在后一部分中选取最大的 $n$ 个元素,使得两部分和的差值最小。 |
| 68 | + |
| 69 | +我们可以用一个大根堆维护前缀中最小的 $n$ 个元素,用一个小根堆维护后缀中最大的 $n$ 个元素。我们定义 $pre[i]$ 表示在数组 $nums$ 的前 $i$ 个元素中选择最小的 $n$ 个元素的和,定义 $suf[i]$ 表示从数组第 $i$ 个元素到最后一个元素中选择最大的 $n$ 个元素的和。在维护大小根堆的过程中,更新 $pre[i]$ 和 $suf[i]$ 的值。 |
| 70 | + |
| 71 | +最后,我们在 $i \in [n, 2n]$ 的范围内枚举分割点,计算 $pre[i] - suf[i + 1]$ 的值,取最小值即可。 |
| 72 | + |
| 73 | +时间复杂度 $O(n \times \log n)$,空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。 |
| 74 | + |
65 | 75 | <!-- tabs:start -->
|
66 | 76 |
|
67 | 77 | ### **Python3**
|
68 | 78 |
|
69 | 79 | <!-- 这里可写当前语言的特殊实现逻辑 -->
|
70 | 80 |
|
71 | 81 | ```python
|
72 |
| - |
| 82 | +class Solution: |
| 83 | + def minimumDifference(self, nums: List[int]) -> int: |
| 84 | + m = len(nums) |
| 85 | + n = m // 3 |
| 86 | + |
| 87 | + s = 0 |
| 88 | + pre = [0] * (m + 1) |
| 89 | + q1 = [] |
| 90 | + for i, x in enumerate(nums[: n * 2], 1): |
| 91 | + s += x |
| 92 | + heappush(q1, -x) |
| 93 | + if len(q1) > n: |
| 94 | + s -= -heappop(q1) |
| 95 | + pre[i] = s |
| 96 | + |
| 97 | + s = 0 |
| 98 | + suf = [0] * (m + 1) |
| 99 | + q2 = [] |
| 100 | + for i in range(m, n, -1): |
| 101 | + x = nums[i - 1] |
| 102 | + s += x |
| 103 | + heappush(q2, x) |
| 104 | + if len(q2) > n: |
| 105 | + s -= heappop(q2) |
| 106 | + suf[i] = s |
| 107 | + |
| 108 | + return min(pre[i] - suf[i + 1] for i in range(n, n * 2 + 1)) |
73 | 109 | ```
|
74 | 110 |
|
75 | 111 | ### **Java**
|
76 | 112 |
|
77 | 113 | <!-- 这里可写当前语言的特殊实现逻辑 -->
|
78 | 114 |
|
79 | 115 | ```java
|
| 116 | +class Solution { |
| 117 | + public long minimumDifference(int[] nums) { |
| 118 | + int m = nums.length; |
| 119 | + int n = m / 3; |
| 120 | + long s = 0; |
| 121 | + long[] pre = new long[m + 1]; |
| 122 | + PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a); |
| 123 | + for (int i = 1; i <= n * 2; ++i) { |
| 124 | + int x = nums[i - 1]; |
| 125 | + s += x; |
| 126 | + pq.offer(x); |
| 127 | + if (pq.size() > n) { |
| 128 | + s -= pq.poll(); |
| 129 | + } |
| 130 | + pre[i] = s; |
| 131 | + } |
| 132 | + s = 0; |
| 133 | + long[] suf = new long[m + 1]; |
| 134 | + pq = new PriorityQueue<>(); |
| 135 | + for (int i = m; i > n; --i) { |
| 136 | + int x = nums[i - 1]; |
| 137 | + s += x; |
| 138 | + pq.offer(x); |
| 139 | + if (pq.size() > n) { |
| 140 | + s -= pq.poll(); |
| 141 | + } |
| 142 | + suf[i] = s; |
| 143 | + } |
| 144 | + long ans = 1L << 60; |
| 145 | + for (int i = n; i <= n * 2; ++i) { |
| 146 | + ans = Math.min(ans, pre[i] - suf[i + 1]); |
| 147 | + } |
| 148 | + return ans; |
| 149 | + } |
| 150 | +} |
| 151 | +``` |
80 | 152 |
|
| 153 | +### **C++** |
| 154 | + |
| 155 | +```cpp |
| 156 | +class Solution { |
| 157 | +public: |
| 158 | + long long minimumDifference(vector<int>& nums) { |
| 159 | + int m = nums.size(); |
| 160 | + int n = m / 3; |
| 161 | + |
| 162 | + using ll = long long; |
| 163 | + ll s = 0; |
| 164 | + ll pre[m + 1]; |
| 165 | + priority_queue<int> q1; |
| 166 | + for (int i = 1; i <= n * 2; ++i) { |
| 167 | + int x = nums[i - 1]; |
| 168 | + s += x; |
| 169 | + q1.push(x); |
| 170 | + if (q1.size() > n) { |
| 171 | + s -= q1.top(); |
| 172 | + q1.pop(); |
| 173 | + } |
| 174 | + pre[i] = s; |
| 175 | + } |
| 176 | + s = 0; |
| 177 | + ll suf[m + 1]; |
| 178 | + priority_queue<int, vector<int>, greater<int>> q2; |
| 179 | + for (int i = m; i > n; --i) { |
| 180 | + int x = nums[i - 1]; |
| 181 | + s += x; |
| 182 | + q2.push(x); |
| 183 | + if (q2.size() > n) { |
| 184 | + s -= q2.top(); |
| 185 | + q2.pop(); |
| 186 | + } |
| 187 | + suf[i] = s; |
| 188 | + } |
| 189 | + ll ans = 1e18; |
| 190 | + for (int i = n; i <= n * 2; ++i) { |
| 191 | + ans = min(ans, pre[i] - suf[i + 1]); |
| 192 | + } |
| 193 | + return ans; |
| 194 | + } |
| 195 | +}; |
| 196 | +``` |
| 197 | + |
| 198 | +### **Go** |
| 199 | + |
| 200 | +```go |
| 201 | +func minimumDifference(nums []int) int64 { |
| 202 | + m := len(nums) |
| 203 | + n := m / 3 |
| 204 | + s := 0 |
| 205 | + pre := make([]int, m+1) |
| 206 | + q1 := hp{} |
| 207 | + for i := 1; i <= n*2; i++ { |
| 208 | + x := nums[i-1] |
| 209 | + s += x |
| 210 | + heap.Push(&q1, -x) |
| 211 | + if q1.Len() > n { |
| 212 | + s -= -heap.Pop(&q1).(int) |
| 213 | + } |
| 214 | + pre[i] = s |
| 215 | + } |
| 216 | + s = 0 |
| 217 | + suf := make([]int, m+1) |
| 218 | + q2 := hp{} |
| 219 | + for i := m; i > n; i-- { |
| 220 | + x := nums[i-1] |
| 221 | + s += x |
| 222 | + heap.Push(&q2, x) |
| 223 | + if q2.Len() > n { |
| 224 | + s -= heap.Pop(&q2).(int) |
| 225 | + } |
| 226 | + suf[i] = s |
| 227 | + } |
| 228 | + ans := int64(1e18) |
| 229 | + for i := n; i <= n*2; i++ { |
| 230 | + ans = min(ans, int64(pre[i]-suf[i+1])) |
| 231 | + } |
| 232 | + return ans |
| 233 | +} |
| 234 | + |
| 235 | +func min(a, b int64) int64 { |
| 236 | + if a < b { |
| 237 | + return a |
| 238 | + } |
| 239 | + return b |
| 240 | +} |
| 241 | + |
| 242 | +type hp struct{ sort.IntSlice } |
| 243 | + |
| 244 | +func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] } |
| 245 | +func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) } |
| 246 | +func (h *hp) Pop() interface{} { |
| 247 | + a := h.IntSlice |
| 248 | + v := a[len(a)-1] |
| 249 | + h.IntSlice = a[:len(a)-1] |
| 250 | + return v |
| 251 | +} |
81 | 252 | ```
|
82 | 253 |
|
83 | 254 | ### **TypeScript**
|
84 | 255 |
|
85 | 256 | ```ts
|
86 |
| - |
| 257 | +function minimumDifference(nums: number[]): number { |
| 258 | + const m = nums.length; |
| 259 | + const n = Math.floor(m / 3); |
| 260 | + let s = 0; |
| 261 | + const pre: number[] = Array(m + 1); |
| 262 | + const q1 = new MaxPriorityQueue(); |
| 263 | + for (let i = 1; i <= n * 2; ++i) { |
| 264 | + const x = nums[i - 1]; |
| 265 | + s += x; |
| 266 | + q1.enqueue(x, x); |
| 267 | + if (q1.size() > n) { |
| 268 | + s -= q1.dequeue().element; |
| 269 | + } |
| 270 | + pre[i] = s; |
| 271 | + } |
| 272 | + s = 0; |
| 273 | + const suf: number[] = Array(m + 1); |
| 274 | + const q2 = new MinPriorityQueue(); |
| 275 | + for (let i = m; i > n; --i) { |
| 276 | + const x = nums[i - 1]; |
| 277 | + s += x; |
| 278 | + q2.enqueue(x, x); |
| 279 | + if (q2.size() > n) { |
| 280 | + s -= q2.dequeue().element; |
| 281 | + } |
| 282 | + suf[i] = s; |
| 283 | + } |
| 284 | + let ans = Number.MAX_SAFE_INTEGER; |
| 285 | + for (let i = n; i <= n * 2; ++i) { |
| 286 | + ans = Math.min(ans, pre[i] - suf[i + 1]); |
| 287 | + } |
| 288 | + return ans; |
| 289 | +} |
87 | 290 | ```
|
88 | 291 |
|
89 | 292 | ### **...**
|
|
0 commit comments