[Leetcode] Search Insert Position

二分查找算法实现

最新推荐文章于 2019-02-13 15:00:16 发布

leo524891010

最新推荐文章于 2019-02-13 15:00:16 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/algorithmengine/article/details/8262989

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class Solution {
public:
    int searchInsert(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int left = 0;
        int right = n;
        
        int lm, rm;
        while (left < right)
        {
            int mid = left + (right - left) / 2;
            
            if (mid == n)
            {
                lm = A[mid - 1];
                rm = INT_MAX;
            }
            else if (mid == 0)
            {
                lm = INT_MIN;
                rm = A[mid];
            }
            else
            {
                lm = A[mid - 1];
                rm = A[mid];
            }
            
            if (target > lm && target <= rm)
                return mid;
            else if (target > rm)
            {
                left = mid + 1;
            }
            else 
            {
                right = mid - 1;
            }
        }
    }
};

class Solution {
public:
    int searchInsert(int a[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (n == 0) return 0;
        
        int left = 0;
        int right = n;
        while (left <= right)
        {
            int mid = left + (right - left) / 2;
            int midE = getValue(a, mid, n);
            int leftE = getValue(a, left, n);
            int rightE = getValue(a, right, n);            
                        
            if (midE >= target && target > getValue(a, mid - 1,n))
            {
                return mid;
            }
            else if (target > midE)
            {
                left = mid + 1;
            }
            else
                right = mid - 1;
        }
    }
    
    int getValue(int a[], int idx, int n)
    {
        if (idx == -1)
            return INT_MIN;
        else if (idx == n)
            return INT_MAX;
        else
            return a[idx];
    }
};