[LeetCode] 034. Search for a Range (Medium) (C++/Java)
2015-03-13 18:34
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索引:[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode
035. Search for a Range (Medium)
代码(github):https://github.com/illuz/leetcode
(C++)直接用 C++ STL 的
(Java)直接从普通的二分改一下就行了。
Java:
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。(C++)直接用 C++ STL 的
lower_bound和
upper_bound偷懒。
(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; } }
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