LeetCode OJ - Search Insert Position

最新推荐文章于 2016-10-10 10:03:57 发布

原创最新推荐文章于 2016-10-10 10:03:57 发布 · 375 阅读

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本文介绍了一个用于在已排序数组中查找目标值的插入位置的算法。通过逐步迭代，实现找到目标值在数组中的正确索引，或者确定其应该插入的位置以保持数组的有序性。该算法考虑了多种边界条件，包括目标值等于数组元素、大于数组最大值或小于数组最小值的情况，并通过二分查找优化了搜索过程。

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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

public class Solution_35 {
    public static int searchInsert(int[] nums, int target) {
    	int pos = -1;
    	int len = nums.length;
    	for(int i=0;i<len;i++){
    		if(target == nums[i])
    			pos = i;
    		else if(target > nums[len-1])
    			pos = len;
    		else if(target < nums[0])
    			pos = 0;
    	}

    	if(pos == -1){
    		int left = 0;
    		int right = len-1;
    		int mid = (left+right)/2;
    		while(nums[left]<=nums[right]){
    			if(target<nums[mid]){
    				right=mid-1;
    				mid = (left+right)/2;
    			}
    			else if(target>nums[mid]){
    				left=mid+1;
    				mid = (left+right)/2;
    			}
    		}
    		System.out.println("left=" + left + ",mid=" + mid + ",right=" + right);
    		pos = mid+1;
    	}
    	return pos;     
    }

    public static void main(String[] args){
    	//int[] nums = {1,2,4,6,7};
    	//int[] nums = {1,3,5,6};
    	int[] nums = {1,3};
    	int num = searchInsert(nums,2);
    	System.out.println(num);
    }
}

虽然 Runtime才1ms 但是只击败了1.7%的人。。唔，明儿再跟别人学学去。