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

標籤：c++   leetcode   演算法   java   

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

035. Search for a Range (Medium)
連結：

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

題意：

在有序數組中找到一個數的範圍。（因為數有重複）

分析：

還是二分搜尋變形。

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

代碼：

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};elsereturn 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)