LeetCode——二分查找

LeetCode——二分查找（时间复杂度O(log n)）

LeetCode Binary Search Summary
二分搜索法小结

1、求开方
T69. Sqrt(x) (Easy)

class Solution {
public:
    int mySqrt(int x) {
        if (x <= 1) return x;
        int left = 0, right = x;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (x / mid >= mid) left = mid + 1;
            else right = mid;
        }
        return right - 1;
        
    }
};

2、大于给定元素的最小元素
T744. Find Smallest Letter Greater Than Target (Easy)

class Solution {
public:
    char nextGreatestLetter(vector<char>& letters, char target) {
        if (target >= letters.back()) return letters[0];
        int n = letters.size(), left = 0, right = n;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (letters[mid] <= target) left = mid + 1;
            else right = mid;
        }
        return letters[right];

    }
};

3、有序数组的 Single Element
T540. Single Element in a Sorted Array (Medium)

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int left = 0, right = nums.size() - 1, n = nums.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == nums[mid + 1]) {
                if ((n - 1 - mid) % 2 == 1) right = mid;
                else left = mid + 1;
            } else {
                if (mid == 0 || nums[mid] != nums[mid - 1]) return nums[mid];
                if ((n - 1 - mid) % 2 == 0) right = mid;
                else left = mid + 1;
            }
        }
        return nums[left];

    }
};

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        int left = 0, right = nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            //int n = mid ^ 1;
            //int n = m % 2 == 0 ? m + 1 : m - 1;
            if (nums[mid] == nums[mid ^ 1]) left = mid + 1;
            else right = mid;
        }
        return nums[left];
    }
};

4、第一个错误的版本
T278. First Bad Version (Easy)

// The API isBadVersion is defined for you.
// bool isBadVersion(int version);

class Solution {
public:
    int firstBadVersion(int n) {
        int left = 1, right = n;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (isBadVersion(mid)) right = mid;
            else left = mid + 1;
        }
        return left;
        
    }
};

5、旋转数组的最小数字
T153. Find Minimum in Rotated Sorted Array (Medium)

class Solution {
public:
    int findMin(vector<int>& nums) {
        int left = 0, right = (int)nums.size() - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > nums[right]) left = mid + 1;
            else right = mid;
        }
        return nums[right];

    }
};

6、查找区间
T34. Find First and Last Position of Element in Sorted Array

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int idx = search(nums, 0, nums.size() - 1, target);
        if (idx == -1) return {-1, -1};
        int left = idx, right = idx;
        while (left > 0 && nums[left - 1] == nums[idx]) --left;
        while (right < nums.size() - 1 && nums[right + 1] == nums[idx]) ++right;
        return {left, right};
    }
    int search(vector<int>& nums, int left, int right, int target) {
        if (left > right) return -1;
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) return mid;
        if (nums[mid] < target) return search(nums, mid + 1, right, target);
        else return search(nums, left, mid - 1, target);
        
    }
};