有序数组中查找
1. LeetCode 34. Find First and Last Position of Element in Sorted Array
1 class Solution {
2 public:
3 vector<int> searchRange(vector<int>& nums, int target) {
4 vector<int> ans = {-1, -1};
5 if(nums.size() == 0) return ans;
6 int l=0, r=nums.size();
7 // 找到最左边的元素
8 while(l < r){
9 int mid = l + (r - l) /2;
10 if(nums[mid] >= target)
11 r = mid;
12 else
13 l = mid + 1;
14 }
15 if(l==nums.size() || nums[l] != target)
16 return ans;
17
18 ans[0] = l;
19 // 找最右边的元素
20 r = nums.size();
21 while(l < r){
22 int mid = l + (r - l) /2;
23 if(nums[mid] > target)
24 r = mid;
25 else
26 l = mid+1;
27 // 错误写法
28 //if(nums[mid] > target) r = mid - 1;
29 //else l = mid; // 当l==mid, r保持不变时会死循环，
30 }
31 ans[1] = l-1;
32 return ans;
33 }
34 };
有序数组的旋转
1 class Solution {
2 public:
3 int search(vector<int>& nums, int target) {
4 if(nums.empty()) return -1;
5 int l=0, h=nums.size()-1;
6 while(l < h){
7 int mid = l + (h - l) / 2;
8 // mid在旋转数组的尾端
9 if(nums[mid] < nums[h])
10 if(target > nums[mid] && target <= nums[h])
11 l= mid + 1;
12 else
13 h = mid;
14 // mid在旋转数组的头端
15 else
16 if(target >= nums[l] && target <= nums[mid])
17 h = mid;
18 else
19 l = mid + 1;
20 }
21 return nums[l]==target ? l : -1;
22 }
23 };
无序数组也可以用二分搜索, 查找局部最大/最小值
1 class Solution {
2 public:
3 int findPeakElement(vector<int>& nums) {
4 if(nums.empty()) return -1;
5 int l=0, h=nums.size()-1;
6 while(l < h){
7 int mid = l + (h - l) / 2;
8 if(nums[mid] < nums[mid+1])
9 l = mid + 1;
10 // 此时nums[mid]>nums[mid+1], 若nums[mid]>nums[mid-1],则nums[mid]
11 // 就是峰值，否则nums[mid]<nums[mid-1],表明峰值在左半区间。画图看一下就明白了
12 else
13 h = mid;
14 }
15 return l;
16 }
17 };
1 class Solution {
2 public:
3 int findDuplicate(vector<int>& nums) {
4 int l=1, n=nums.size(), h=n-1;
5 // l,h不是索引，是数的取值范围
6 while(l < h){
7 int mid = l + (h - l) / 2; // 中间大小的数
8 int cnt=0; // 计算比mid小的数有几个
9 for(int i=0; i<n; ++i)
10 if(nums[i] <= mid)
11 ++cnt;
12 // 比mid小的数个数比mid大，表明1~mid中间有重复的数字
13 if(cnt > mid)
14 h = mid;
15 else
16 l = mid + 1;
17 }
18 return l;
19 }
20 };