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Q:
A: binary search, pay attention to boundary cases
java version 1
public class Solution { public int searchInsert(int[] nums, int target) { if (nums == null || nums.length == 0) return 0; int start = 0; int end = nums.length-1; while(start +1 < end){ int mid = start + (end - start) / 2; if (nums[mid] == target){ return mid; }else if (nums[mid] < target){ start = mid; }else{ end = mid; } } if (nums[start] >= target) return start; else if (target <= nums[end]) return end; else return end+1; } }
Leetcode 35 Search insert position
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原文地址:http://www.cnblogs.com/nobody2somebody/p/5146216.html