[LeetCode] 034. Search for a Range (Medium) (C++/Java)

标签：c++   leetcode   算法   java   

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode

035. Search for a Range (Medium)

链接：

题目：https://leetcode.com/problems/search-for-a-range/
代码(github)：https://github.com/illuz/leetcode

题意：

在有序数组中找到一个数的范围。（因为数有重复）

分析：

还是二分搜索变形。

1. (C++)直接用 C++ STL 的 lower_bound 和 upper_bound 偷懒。
2. (Java)直接从普通的二分改一下就行了。

代码：

C++:

class Solution {
public:
	vector<int> searchRange(int A[], int n, int target) {
		int* lower = lower_bound(A, A + n, target);
		int* upper = upper_bound(A, A + n, target);
		if (*lower != target)
			return vector<int> {-1, -1};
		else
			return vector<int>{lower - A, upper - A - 1};
	}
};

Java:

public class Solution {

    public int[] searchRange(int[] A, int target) {
        int[] ret = new int[2];
        ret[0] = ret[1] = -1;
        int left = 0, right = A.length - 1, mid;

        while (left <= right) {
            if (A[left] == target && A[right] == target) {
                ret[0] = left;
                ret[1] = right;
                break;
            }

            mid = (right + left) / 2;
            if (A[mid] < target) {
                left = mid + 1;
            } else if (A[mid] > target) {
                right = mid - 1;
            } else {
                if (A[right] == target) {
                    ++left;
                } else {
                    --right;
                }
            }
        }

        return ret;
    }
}

[LeetCode] 034. Search for a Range (Medium) (C++/Java)

标签：c++   leetcode   算法   java   

原文地址：http://blog.csdn.net/hcbbt/article/details/44244631