【leetcode】Search in Rotated Sorted Array II（middle）☆

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

我的思路：

太混乱了不提了。注意关键区分依据排好序的一定是从小到大的

看大神的吧：

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 Iif (A[l] < A[m]) { //left half is sorted 排好序的部分一定是从小到大的if (A[l] <= key && key < A[m]) //在排好序的这部分r = m - 1;elsel = m + 1;} else if (A[l] > A[m]) { //right half is sortedif (A[m] < key && key <= A[r])l = m + 1;elser = m - 1;} else l++; //A[l] == A[m] 把l增大1个再循环
    }return false;
}

我的代码，把三个数字都相等的情况单独处理，其他就用无重复处理。其实我的代码看起来长一些，但是在处理1111111111115这种情况时我的方法优势还是有的

bool search(int A[], int n, int target) {int l = 0, r = n - 1;while(l <= r){int m = (l + r) / 2;if(target == A[m])return true;else if(A[l] == A[m] && A[m] == A[r]) //三个值相等
        {bool isequalleft = true;for(int i = l; i < m; i++){if(A[i] != A[i + 1]){isequalleft = false;break;}}if(isequalleft) //去掉重复的那一半l = m + 1;elser = m - 1;}else //与不重复的方法相同
        {if (A[l] <= A[m]) {if (target >= A[l] && target < A[m]) {r = m - 1;} else {l = m + 1;}} else {if (target > A[m] && target <= A[r]) {l = m + 1;} else {r = m - 1;}}}}return false;
}

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【leetcode】Search in Rotated Sorted Array II（middle）☆

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