Search for a Range

Search for a Range

问题：

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

思路：

　　递归二分查找问题

我的代码：

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int[] rst = {-1,-1};
        if(A == null || A.length == 0) return rst;
        int left = 0;
        int right = A.length - 1;
        return helper(A, target, left, right);
    }
    public int[] helper(int[] A, int target, int left, int right)
    {
        int[] rst = {-1,-1};
        if(left > right) return rst;
        if(left == right)
        {
            if(target == A[left])
            {
                rst[0] = left;
                rst[1] = left;
                return rst;
            }
            return rst;
        }
        int mid = (left + right)/2;
        if(A[mid] == target)
        {
            rst[0] = mid;
            rst[1] = mid;
            int[] leftRange = helper(A, target, left, mid - 1);
            int[] rightRange = helper(A, target, mid + 1, right);
            return mergeRange(leftRange, rightRange, rst);
        }
        else if(A[mid] > target)
            return helper(A, target, left, mid - 1);
        else
            return helper(A, target, mid + 1, right);
        
    }
    public int[] mergeRange(int[] left, int[]right, int[] mid)
    {
        int[] rst = new int[2];
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for(int num: left)
        {
            if(num >= 0)
            {
                min = Math.min(min, num);
                max = Math.max(max, num);
            }
        }
        for(int num: right)
        {
            if(num >= 0)
            {
                min = Math.min(min, num);
                max = Math.max(max, num);
            }
        }
        for(int num: mid)
        {
            if(num >= 0)
            {
                min = Math.min(min, num);
                max = Math.max(max, num);
            }
        }
        rst[0] = min;
        rst[1] = max;
        return rst;
    }
    
}

他人代码：

public class Solution {
    public int[] searchRange(int[] A, int target) {
        int start, end, mid;
        int[] bound = new int[2]; 
        
        // search for left bound
        start = 0; 
        end = A.length - 1;
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (A[mid] == target) {
                end = mid;
            } else if (A[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (A[start] == target) {
            bound[0] = start;
        } else if (A[end] == target) {
            bound[0] = end;
        } else {
            bound[0] = bound[1] = -1;
            return bound;
        }
        
        // search for right bound
        start = 0;
        end = A.length - 1;
        while (start + 1 < end) {
            mid = start + (end - start) / 2;
            if (A[mid] == target) {
                start = mid;
            } else if (A[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (A[end] == target) {
            bound[1] = end;
        } else if (A[start] == target) {
            bound[1] = start;
        } else {
            bound[0] = bound[1] = -1;
            return bound;
        }
        
        return bound;
    }
}

学习之处：

• 自己的代码虽然AC了，但是判断太多了，一点也不简洁，这为以后的调试，改代码带来了痛苦，而且代码越多越容易出bug啊
• 看看别人的代码，被人代码的模板之前也整理过，但是就没想到用呢，首先left + 1 <right 这种判断的存在是因为 之后的二分存在 left = mid or right = mid 而不是left = mid - 1 right = mid + 1 如果不是Left + 1 < right 则存在死循环
• 对于不知道左边还是右边的：首先考虑可不可以排除一边，如通过left mid right值三者之间的大小关系，如果不能排除，更加普世的作用是使用递归分别dfsleft dfsright进行结合，其实也可以在循环里面做了，方法是首先判断左边，然后判断右边，代码就如同上面别人的代码一样，真实巧妙！