昨天看到網(wǎng)上有人說leetcode不能提交成功就好，要想運行時間短的算法。今天有點時間就對前面的題想想吧。主要還是看別人的，但是提交之后發(fā)現(xiàn)和我的差別不是很大，1ms。
34題Java代碼，思路是用二分法。二分法思想很簡單，就不多說了。代碼轉自http://blog.csdn.net/lilong_dream/article/details/22893675

public int[] searchRange(int[] A, int target) {
        int left = 0;
        int right = A.length - 1;
        int[] result = { -1, -1 };
        while (left <= right) {
            int mid = (left + right) / 2;
            if (A[mid] > target) {
                right = mid - 1;
            } else if (A[mid] < target) {
                left = mid + 1;
            } else {
                result[0] = mid;
                result[1] = mid;
                int i = mid - 1;
                while (i >= 0 && A[i] == target) {
                    result[0] = i;
                    --i;
                }
                i = mid + 1;
                while (i < A.length && A[i] == target) {
                    result[1] = i;
                    ++i;
                }
                break;
            }
        }
        return result;
    }

提交之后大概擊敗26%，比我的快了1ms。

35題還是利用二分法。Java代碼轉自http://blog.csdn.net/u012848330/article/details/52145150

public class Solution {
    public int searchInsert(int[] nums, int target) {
        return binarySearch(nums,target,0,nums.length-1);
    }
    private int binarySearch(int[] nums, int target, int left, int right) {
        // TODO Auto-generated method stub
        if(left > right)
            return -1;
        if(left == right){
            if(nums[left] >= target){
                return left;
            }else{
                return right+1;
            }
        }
        int mid = (left + right)/2;
        if(target < nums[mid]){
            return binarySearch(nums,target,left,mid);
        }else if(target == nums[mid]){
            return mid;
        }else{
            return binarySearch(nums,target,mid+1,right);
        }
    }
}

提交之后還是比我的快了1ms。這個思路算是不錯的了。兩個題都是二分法。