局部最小点

http://haixiaoyang.wordpress.com/category/search-binary-search/

/*Suppose we are given an array A[1 .. n] with the special property that A[1] ≥A[2] andA[n − 1] ≤A[n]. We say that an element A[x] is a local minimum if it is less than or equalto both its neighbors, or more formally, if A[x − 1] ≥A[x] and A[x] ≤A[x + 1]. For example,there are five local minima in the following array:9 7 7 2 1 3 7 5 4 7 3 3 4 8 6 9We can obviously find a local minimum in O(n) time by scanning through the array. Describeand analyze an algorithm that finds a local minimum in O(log n) time*///Same as previous binary search, based on location! When if returns?//When it should search left, when it should search right?int findLocalMin(int a[], int n){assert(a && n > 2);int nBeg = 0;int nEnd = n-1;while (nBeg <= nEnd){int nMid = nBeg + (nEnd - nBeg)/2;//This logic is straight forwardif (nMid > 0 && nMid < n-1 && a[nMid-1] >= a[nMid] && a[nMid+1] >= a[nMid])return nMid;//The most important thing is how to come up with the logic below://1: should search left if in the right most point, obvious//2: UNDER THE SITUATION THAT nMid is not a middle point,//   can judge the trend by (nMid, nMid+1) or (nMid-1, nMid)//   take advantage of nMid != n-1 when reaching the right half of logicif ((nMid == n-1) || (a[nMid] <= a[nMid+1])) //Pitfall, less equal than not less!!nEnd = nMid - 1;else nBeg = nMid + 1; //Otherwise logic}return -1;}

一定要注意这个条件，a[1] >=a[2]    a[n-1] <= a[n]

那么在此区间内肯定至少有一个局部最小点

如果  a[mid-1]<=a[mid] <= a[mid+1]

则在beg-mid区间肯定有局部最小

如果a[mid-1]>= a[mid]>=a[mid+1]

则在mid-end区间肯定有局部最小

如果 a[mid-1]<= a[mid]>=a[mid+1]

在两个区间内均存在局部最小