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Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Here are few examples.[1,3,5,6], 5 → 2[1,3,5,6], 2 → 1[1,3,5,6], 7 → 4[1,3,5,6], 0 → 0
方法一：线性搜索
class Solution { public: int searchInsert(int A[], int n, int target) { int i; for( i = 0; i < n; i++) { if(A[i] >= target) return i; } return i; } };
方法二：此题的意图应该不是线性搜索，应该是二分
注意两点，第一，当A[mid] > target时，mid本身有可能就是答案，
第二，当left==right时，由于 target有可能大于A[right],这时，要返回right+1，而不是right
class Solution { public: #if 0 int searchInsert(int A[], int n, int target) { int i; for( i = 0; i < n; i++) { if(A[i] >= target) return i; } return i; } #endif int searchInsert(int A[], int n, int target) { return binarySearch(A, 0, n-1, target); } int binarySearch(int A[], int left, int right, int target) { if(left == right) { if(A[left] >= target) return left; else return right + 1; } int mid = (left + right)/2; if(A[mid] == target) return mid; if(A[mid] > target)//mid may be the answer return binarySearch(A, left, mid , target); if(A[mid] < target) return binarySearch(A, mid+1, right, target); } };
[LeetCode] Search Insert Position
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原文地址：http://www.cnblogs.com/diegodu/p/4283271.html
