[leetcode]Search in Rotated Sorted Array II
这道题目仍然是二分,去掉不可能的部分。用了递归,在重复的情况下,就是有可能最左边的和最右边的相等,此时就不能直接判断出区间外的元素,左右两边同时递归。有重复元素的时候,在bad case的时候会退化为O(n)
public class Solution { public boolean search(int[] A, int target) { return search(A, target, 0, A.length-1); } private boolean search(int[] A, int target, int left, int right) { if( left > right) return false; if (left == right) return (A[left] == target); if (A[left] < A[right] && (target < A[left] || target > A[right])) return false; if (A[left] > A[right] && target > A[right] && target < A[left]) return false; int mid = left + (right - left) / 2; if (A[mid] == target) return true; else { return search(A, target, left, mid-1) || search(A, target, mid+1, right); } } }
网上有个思路更简洁,二分。如果相等导致不能二分,则直接跳过这个重复的元素。
class Solution { public: bool search(int A[], int n, int key) { int l = 0, r = n - 1; while (l <= r) { int m = l + (r - l)/2; if (A[m] == key) return true; //return m in Search in Rotated Array I if (A[l] < A[m]) { //left half is sorted if (A[l] <= key && key < A[m]) r = m - 1; else l = m + 1; } else if (A[l] > A[m]) { //right half is sorted if (A[m] < key && key <= A[r]) l = m + 1; else r = m - 1; } else l++; } return false; } };