【算法】【二分】Leetcode二分问题总结

在排序数组中查找数字

题目链接：https://leetcode-cn.com/problems/zai-pai-xu-shu-zu-zhong-cha-zhao-shu-zi-lcof/

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size() - 1;
        int l = 0;
        int r = n;
        while(l < r)
        {
            int mid = l + r >> 1;
            if(nums[mid] >= target) r = mid;
            else l = mid + 1;
        }
        cout << r << endl;
        int res = 0;
        int i = l;
        while(i <= n && nums[i++] == target)
            res++;
        return res;
    }
};

搜索旋转排序数组

题目链接：https://leetcode-cn.com/problems/search-in-rotated-sorted-array/

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        if(n == 0) return -1;

        int l = 0, r = n - 1;
        while(l < r)
        {
            int mid = (l + r + 1) >> 1;
            if(nums[mid] >= nums[0])
                l = mid;
            else
                r = mid - 1;
        }
   
        if (target >= nums[0]){
            // left
            l = 0;
        } else {
            // right
            l = r + 1;
            r = n - 1;
        }

        while(l < r)
        {
            int mid = (l + r) >> 1;
            if(nums[mid] >= target) r = mid;
            else l = mid + 1;
        }
        return nums[r] == target ? r : -1;
    }
};

在排序数组中查找元素的第一个和最后一个位置

题目链接：https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        if(nums.size() == 0) return {-1, -1};
        vector<int> res(2, -1);
        res[0] = left(nums, target);
        res[1] = right(nums, target);
    
        return res;
    }
    int left(vector<int>& nums, int target)
    {
        int n = nums.size();
        int l = 0, r = n - 1;
        while(l < r)
        {
            int mid = (l + r)  >> 1;
            if(nums[mid] >= target) r = mid;
            else l = mid + 1;
        }
        return target != nums[l] ? -1 : l;
    }

    int right(vector<int>& nums, int target)
    {
        int n = nums.size();
        int l = 0, r = n - 1;
        while(l < r)
        {
            int mid = (l + r + 1)  >> 1;
            if(nums[mid] <= target) l = mid;
            else r = mid - 1;
        }
        return target != nums[l] ? -1 : l;
    }
};

寻找两个正序数组的中位数

题目链接:https://leetcode-cn.com/problems/median-of-two-sorted-arrays/

归并

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        vector<int> nums;
        int i, j;
        for(i = 0, j = 0; i < nums1.size() && j < nums2.size(); )
        {
            if(nums1[i] < nums2[j]) nums.push_back(nums1[i++]);
            else nums.push_back(nums2[j++]);
        }
        while(i < nums1.size()) nums.push_back(nums1[i++]);
        while(j < nums2.size()) nums.push_back(nums2[j++]);

        //for_each(nums.begin(), nums.end(), [](int x) {cout << x << " ";});
        cout << endl;
        int n = nums.size();
        if(n & 1) return nums[(n >> 1)];
        else return (nums[(n >> 1) - 1] + nums[(n >> 1)]) / 2.0;
    }
};

用二分求两个有序数组第k个数

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int len = nums1.size() + nums2.size();
        //两个数组长度和为奇数
        if( len & 1) return getKth(nums1, nums2, (len >> 1) + 1);
        //长度和为偶数
        else return (getKth(nums1, nums2, len >> 1) + getKth(nums1, nums2, (len >> 1) + 1)) / 2.0;
    }

    int getKth(vector<int>& nums1, vector<int>& nums2, int k)
    {
            int n = nums1.size(), m = nums2.size();
            int l = max(0, k - m), r = min(k, n);

            //循环结束，l即为所求位置；
            //第k小即为max(nums1[l - 1], num2(k - l - 1));
          
            while(l < r)
            {
                int mid = (l + r ) >> 1;
                if(nums2[k - mid - 1] > nums1[mid]) l = mid + 1;
                else r = mid; 
            }
            //由于l可以为0 和 k；
            //所以l - 1和k - l - 1可能不存在，需要单独判断，防止越界
            int n1_l_max = (l == 0) ? INT_MIN:nums1[l - 1];
            int n2_2_max = (l == k) ? INT_MIN:nums2[k - l - 1];
            return max(n1_l_max, n2_2_max);
    }
};

三数之和

题目链接：https://leetcode-cn.com/problems/3sum/

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        if(nums.size() < 3) return {};
        sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        for(int i = 0; i < nums.size() - 2; ++i)
        {
            if(nums[i] > 0) break;
            if(i > 0 && nums[i] == nums[i - 1])  continue;

            int target = -nums[i];
            int l = i + 1, r = nums.size() - 1;
            while(l < r)
            {
                int sum = nums[l] + nums[r];
                if(sum == target)
                {
                    res.push_back({nums[i], nums[l], nums[r]});
                    l++;
                    r--;
                    while(l < r && nums[l] == nums[l - 1]) l++;
                    while(l < r && nums[r] == nums[r + 1]) r--;
                }
                if(sum < target) l++;
                if(sum > target) r--;
            }
        }
        return res;
    }
};