|
| 1 | +## List-binarysearch |
| 2 | +### 二分查找 |
| 3 | +#### 解题步骤 |
| 4 | +1. 确定二分左右边界 |
| 5 | +2. 设计一个性质使得可以将区间划分成两段(要求的点在其中一段的端点) |
| 6 | +3. 区间更新 |
| 7 | +#### 二分查找模板 |
| 8 | +``` |
| 9 | +bool check(int x) // 检查x是否满足某种性质 |
| 10 | +int bsearch_1(int l, int r){ |
| 11 | + while (l < r){ |
| 12 | + int mid = l + r >> 1; |
| 13 | + if (check(mid)) r = mid; |
| 14 | + else l = mid + 1; |
| 15 | + } |
| 16 | + return l; |
| 17 | +} |
| 18 | + |
| 19 | +int bsearch_2(int l, int r){ |
| 20 | + while (l < r){ |
| 21 | + int mid = l + r + 1>> 1; |
| 22 | + if (check(mid)) l = mid; |
| 23 | + else r = mid - 1; |
| 24 | + } |
| 25 | + return l; |
| 26 | +} |
| 27 | +``` |
| 28 | +#### leetcode.704 |
| 29 | +- 链接<https://leetcode.cn/problems/binary-search/> |
| 30 | +- leetcode解题代码 |
| 31 | +``` |
| 32 | +class Solution { |
| 33 | +public: |
| 34 | + int search(vector<int>& nums, int target) { |
| 35 | + int l = 0, r = nums.size() - 1; |
| 36 | + while (l < r){ |
| 37 | + int mid = (l + r) / 2; |
| 38 | + if (nums[mid] >= target) r = mid; |
| 39 | + else l = mid + 1; |
| 40 | + } |
| 41 | + if (nums[l] == target) return l; |
| 42 | + return -1; |
| 43 | + } |
| 44 | +}; |
| 45 | +``` |
| 46 | +#### leetcode.278 |
| 47 | +- 链接<https://leetcode.cn/problems/first-bad-version/> |
| 48 | +- leetcode解题代码 |
| 49 | +``` |
| 50 | +// The API isBadVersion is defined for you. |
| 51 | +// bool isBadVersion(int version); |
| 52 | + |
| 53 | +class Solution { |
| 54 | +public: |
| 55 | + int firstBadVersion(int n) { |
| 56 | + long l = 1, r = n; |
| 57 | + while (l < r){ |
| 58 | + int mid = (l + r) / 2; |
| 59 | + if (isBadVersion(mid)) r = mid; |
| 60 | + else l = mid + 1; |
| 61 | + } |
| 62 | + return l;// 返回的是第一个错误的版本 |
| 63 | + } |
| 64 | +}; |
| 65 | +``` |
| 66 | +#### leetcode.35 |
| 67 | +- 链接<https://leetcode.cn/problems/search-insert-position/> |
| 68 | +- leetcode解题代码 |
| 69 | +``` |
| 70 | +class Solution { |
| 71 | +public: |
| 72 | + int searchInsert(vector<int>& nums, int target) { |
| 73 | + int n = nums.size(); |
| 74 | + if (target > nums[n - 1]) return n; |
| 75 | + int l = 0, r = n - 1; |
| 76 | + while (l < r){ |
| 77 | + int mid = (l + r) / 2; |
| 78 | + if (nums[mid] >= target) r = mid; |
| 79 | + else l = mid + 1; |
| 80 | + } |
| 81 | + return l; |
| 82 | + } |
| 83 | +}; |
| 84 | +``` |
| 85 | +#### leetcode.69 |
| 86 | +- 链接<https://leetcode.cn/problems/sqrtx/submissions/> |
| 87 | +- leetcode解题代码 |
| 88 | +``` |
| 89 | +class Solution { |
| 90 | +public: |
| 91 | + int mySqrt(int x) { |
| 92 | + if (x <= 1) return x; |
| 93 | + long l = 1, r = x; |
| 94 | + while (l < r){ |
| 95 | + int mid = l + r + 1 >> 1; |
| 96 | + if (mid <= x / mid) l = mid;// 下取整,取第一个小于等于根号下x的值 |
| 97 | + else r = mid - 1; |
| 98 | + } |
| 99 | + |
| 100 | + return l; |
| 101 | + } |
| 102 | +}; |
| 103 | +``` |
| 104 | +#### leetcode.367 |
| 105 | +- 链接<https://leetcode.cn/problems/valid-perfect-square/> |
| 106 | +- 注意:与上一题区别在于只是判断是否存在完全平方数,不需要上取整或者下取整求完全平方数,所以使用模板一或者模板二均可 |
| 107 | +- leetcode解题代码 |
| 108 | +``` |
| 109 | +class Solution { |
| 110 | +public: |
| 111 | + bool isPerfectSquare(int num) { |
| 112 | + long l = 1, r = num; |
| 113 | + while (l < r){ |
| 114 | + int mid = l + r + 1 >> 1; |
| 115 | + if (mid <= num / mid) l = mid; |
| 116 | + else r = mid - 1; |
| 117 | + } |
| 118 | + return r * r == num; |
| 119 | + } |
| 120 | +}; |
| 121 | +``` |
| 122 | +#### leetcode.34 |
| 123 | +- 链接<https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/> |
| 124 | +- leetcode解题代码 |
| 125 | +``` |
| 126 | +class Solution { |
| 127 | +public: |
| 128 | + vector<int> searchRange(vector<int>& nums, int target) { |
| 129 | + if (nums.empty()) return {-1, -1}; |
| 130 | + int l = 0, r = nums.size() - 1; |
| 131 | + while (l < r){ |
| 132 | + int mid = l + r >> 1; |
| 133 | + // 找到>=target的第一个数 |
| 134 | + if (nums[mid] >= target) r = mid; |
| 135 | + else l = mid + 1; |
| 136 | + } |
| 137 | + |
| 138 | + if (nums[l] != target) return {-1, -1}; |
| 139 | + int L = l; |
| 140 | + |
| 141 | + l = 0, r = nums.size() - 1; |
| 142 | + while (l < r){ |
| 143 | + int mid = l + r + 1 >> 1; |
| 144 | + // 找到<=target的第一个数 |
| 145 | + if (nums[mid] <= target) l = mid; |
| 146 | + else r = mid - 1; |
| 147 | + } |
| 148 | + return {L, r}; |
| 149 | + } |
| 150 | +}; |
| 151 | +``` |
| 152 | +#### leetcode.74 |
| 153 | +- 链接<https://leetcode.cn/problems/search-a-2d-matrix/> |
| 154 | +- 解题思路:将二维矩阵展成一维数组,二分 |
| 155 | +- 注意:矩阵下标(i,j)转化为一维数组下标 |
| 156 | +- leetcode解题代码 |
| 157 | +``` |
| 158 | +class Solution { |
| 159 | +public: |
| 160 | + bool searchMatrix(vector<vector<int>>& matrix, int target) { |
| 161 | + if (matrix.empty()) return false; |
| 162 | + int n = matrix.size(), m = matrix[0].size(); |
| 163 | + int l = 0, r = n * m - 1; |
| 164 | + while (l < r){ |
| 165 | + int mid = l + r >> 1; |
| 166 | + // 将矩阵下标转换为数组下标 |
| 167 | + if (matrix[mid / m][mid % m] >= target) r = mid; |
| 168 | + else l = mid + 1; |
| 169 | + } |
| 170 | + |
| 171 | + if (matrix[l / m][l % m] == target) return true; |
| 172 | + return false; |
| 173 | + } |
| 174 | +}; |
| 175 | +``` |
| 176 | +#### leetcode.153 |
| 177 | +- 链接<https://leetcode.cn/problems/find-minimum-in-rotated-sorted-array/> |
| 178 | +- 解题思路:找到数组的二段性(前段的值均大于数组的最后一位,后段的值均小于等于数组的最后一位,并且最小的点在后段的端点处) |
| 179 | +- leetcode解题代码 |
| 180 | +``` |
| 181 | +class Solution { |
| 182 | +public: |
| 183 | + int findMin(vector<int>& nums) { |
| 184 | + int l = 0, r = nums.size(); |
| 185 | + while (l < r){ |
| 186 | + int mid = l + r >> 1; |
| 187 | + if (nums[mid] <= nums.back()) r = mid; |
| 188 | + else l = mid + 1; |
| 189 | + } |
| 190 | + return nums[l]; |
| 191 | + } |
| 192 | +}; |
| 193 | +``` |
| 194 | +#### leetcode.33 |
| 195 | +- 链接<https://leetcode.cn/problems/search-in-rotated-sorted-array/> |
| 196 | +- 解题思路:由于没有二段性不能直接二分,先找到数组的最小值(参考上一题),找到目标值所在区间,在对应区间内二分找目标值 |
| 197 | +- leetcode解题代码 |
| 198 | +``` |
| 199 | +class Solution { |
| 200 | +public: |
| 201 | + int search(vector<int>& nums, int target) { |
| 202 | + if (nums.empty()) return -1; |
| 203 | + // 找到最小值 |
| 204 | + int l = 0, r = nums.size() - 1; |
| 205 | + while (l < r){ |
| 206 | + int mid = (l + r) / 2; |
| 207 | + if (nums[mid] <= nums.back()) r = mid; |
| 208 | + else l = mid + 1; |
| 209 | + } |
| 210 | + // 判断目标值所在区间 |
| 211 | + if (target <= nums.back()) r = nums.size() - 1; |
| 212 | + else l = 0, r --; |
| 213 | + // 在所在区间找目标值 |
| 214 | + while (l < r){ |
| 215 | + int mid = (l + r) / 2; |
| 216 | + if (nums[mid] >= target) r = mid; |
| 217 | + else l = mid + 1; |
| 218 | + } |
| 219 | + if (nums[l] == target) return l; |
| 220 | + return -1; |
| 221 | + } |
| 222 | +}; |
| 223 | +``` |
| 224 | +*** |
| 225 | +下列题目没有二段性,但仍然可以通过二分每次缩小一般的搜索区间找到答案 |
| 226 | +#### leetcode.162 |
| 227 | +- 链接<https://leetcode.cn/problems/find-peak-element/> |
| 228 | +- 解题思路:如果二分的中点的值比中点右边的值小,那么右边一定存在峰值,同理,如果二分的中点的值比中点右边的值大,那么左边一定存在峰值 |
| 229 | +- leetcode解题代码 |
| 230 | +``` |
| 231 | +class Solution { |
| 232 | +public: |
| 233 | + int findPeakElement(vector<int>& nums) { |
| 234 | + int l = 0, r = nums.size() - 1; |
| 235 | + while (l < r){ |
| 236 | + int mid = l + r >> 1; |
| 237 | + // 说明左边一定有峰值 |
| 238 | + if (nums[mid] > nums[mid + 1]) r = mid; |
| 239 | + else l = mid + 1; |
| 240 | + } |
| 241 | + return l; |
| 242 | + } |
| 243 | +}; |
| 244 | + |
| 245 | +class Solution { |
| 246 | +public: |
| 247 | + int findPeakElement(vector<int>& nums) { |
| 248 | + int l = 0, r = nums.size() - 1; |
| 249 | + while (l < r){ |
| 250 | + int mid = l + r >> 1; |
| 251 | + // 说明右边一定有峰值 |
| 252 | + if (nums[mid] < nums[mid + 1]) l = mid + 1; |
| 253 | + else r = mid; |
| 254 | + } |
| 255 | + return l; |
| 256 | + } |
| 257 | +}; |
| 258 | +``` |
| 259 | +#### leetcode.287 |
| 260 | +- 链接<https://leetcode.cn/problems/find-the-duplicate-number/> |
| 261 | +- 解题思路:抽屉原理,n+1个苹果放进n个抽屉中必然有一个抽屉放了两个苹果,抽屉为数据范围n,苹果为数字总数n+1,对数据范围二分,记录二分中点左侧的数字总数,和左边数据范围对比,如果数字总数大于数据范围,说明重复数在左侧,右侧同理 |
| 262 | +- leetcode解题代码 |
| 263 | +``` |
| 264 | +class Solution { |
| 265 | +public: |
| 266 | + int findDuplicate(vector<int>& nums) { |
| 267 | + int l = 1, r = nums.size(); |
| 268 | + while (l < r){ |
| 269 | + int mid = l + r >> 1; |
| 270 | + |
| 271 | + int cnt = 0; |
| 272 | + for (auto c: nums){ |
| 273 | + if (c >= l && c <= mid) |
| 274 | + cnt ++; |
| 275 | + } |
| 276 | + if (cnt > mid - l + 1) r = mid; |
| 277 | + else l = mid + 1; |
| 278 | + } |
| 279 | + return l; |
| 280 | + } |
| 281 | +}; |
| 282 | +``` |
| 283 | +解题参考:<https://www.acwing.com/> |
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