[剑指offer] 数字在排序数组当中出现的次数

[剑指offer] 数字在排序数组当中出现的次数 P221

11122222333

使用二分搜索查找target的开始indexbegin和结束indexend，则个数是indexend - indexbegin + 1;

找到开始位置

getBeginIndex(int [] num, int target):
　　int left =0;

　　int right = num.length-1;

　　while(left<=right){

　　　　if(left<right) return -1;没有这个元素target

　　　　int mid = (left + right)/2;

　　　　if(num[mid]==target){

　　　　　　　if(mid==0||num[mid-]!=target) return mid;

　　　　　　　right = mid-1;

　　　　}else if(num[mid]> target){

　　　　　　　right = mid-1;

　　　　}else{

　　　　　　　left = mid+1;

　　　　}

　　}

return -1;

找到结束位置

getEndIndex(int [] num ,int target):

　　int left = 0;

　　int right = num.length-1;

　　while(left<=right):

　　　　if(left>right) return -1;

　　　　int mid = (right+left)/2;

　　　　if(num[mid]==target){

　　　　　　 if(mid==num.length-1||num[mid+1]!=target) return mid;

　　　　　　left = mid+1;

　　　　}else if(num[mid]>target){

　　　　　　　right = mid-1;

　　　　}else{

　　　　　　　left = mid+1;

　　　　}

return -1;

调用上面的方法求左右边界

targetlength（int [] num,int target）：

　　int left = getBeginIndex(num,target);

　　int right = getEndIndex(num,target);

　　if(left!=-1&&right!=-1):

　　　　return[0]=left;

　　　　return[1]=right;

　　else

　　　　return error;

end

 

上面两个方法都是O（logN）的，所以总的复杂度是O（logN）

 leetcode 当中的一个题目使用这里的方法

search for a range

public class Solution {
    private int searchL(int [] nums,int target){
        //to reduce the thinking and make the problem clear, the corner case should be 
        // resolved first.
        if(target<nums[0]||target>nums[nums.length-1]){
            return -1;
        }
        int left = 0;
        int right = nums.length-1;
        while(left<=right){
            int mid = (left+right)/2;
            if(nums[mid]==target){
                if(mid==0||nums[mid-1]!=target) return mid;
                right = mid-1;
            }else if(nums[mid]>target){
                right = mid-1;
            }else{
                left = mid+1;
            }
        }
        return -1;
    }
    private int searchR(int [] nums,int target){
        //to reduce the thinking and make the problem clear, the corner case should be 
        // resolved first.
        if(target<nums[0]||target>nums[nums.length-1]){
            return -1;
        }
        int left = 0;
        int right = nums.length-1;
        while(left<=right){
            int mid = (left+right)/2;
            if(nums[mid]==target){
                if(mid==nums.length-1||nums[mid+1]!=target) return mid;
                left = mid+1;
            }else if(nums[mid]>target){
                right = mid-1;
            }else{
                left = mid+1;
            }
        }
        //when stop the right index must be larger than the right index
        //the left index can not stop on the index whose value equals target
        //so we just need to consider right index here
        return -1;
    }
    public int[] searchRange(int[] nums, int target) {
        int [] res = new int[2];
        res [0] = searchL(nums,target);
        res [1] = searchR(nums,target);
        return res;
    }
}