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] and A[n ? 1] ≤A[n]. We say that an element A[x] is a local minimum if it is less than or equal to both its neighbors, or more formally, if A[x ? 1] ≥A[x] and A[x] ≤A[x + 1]. For example, there are ?ve local minima in the following array: 9 7 7 2 1 3 7 5 4 7 3 3 4 8 6 9 We can obviously ?nd a local minimum in O(n) time by scanning through the array. Describe and analyze an algorithm that ?nds 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; }

一定要注意這個條件，a[1] >=a[2] ? ?a[n-1] <= a[n]

那么在此區間內肯定至少有一個局部最小點

如果 ?a[mid-1]<=a[mid] <= a[mid+1]

則在beg-mid區間肯定有局部最小

如果a[mid-1]>= a[mid]>=a[mid+1]

則在mid-end區間肯定有局部最小

如果 a[mid-1]<= a[mid]>=a[mid+1]

在兩個區間內均存在局部最小

總結

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