lintcode -- 428. x的n次幂、1319. Contains Duplicate II、156. 合并区间 -- 修改思路

题目

1. 实现 pow(x,n)
2. Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.
3. 给出若干闭合区间，合并所有重叠的部分

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实现代码

• 首先找到小于n的最大的二进制数，这个最大二进制数就是实现对数复杂度的关键
• myPow(x,index / 2)*myPow(x,index / 2) * myPow(x,n-index);中myPow(x,index / 2)*myPow(x,index / 2) = myPow(x,index)应该知道这里使用乘法相当于myPow(x,n)之间n值直接相加
• 使用递归的方法，直到n的值变成0、1、-1.

public class Solution {
    /*
     * @param x: the base number
     * @param n: the power number
     * @return: the result
     */
    public double myPow(double x, int n) {
        // write your code here
        if (n == 0) return 1;
        if (n == 1) return x;
        int index = getMaxTwo(n);
        return myPow(x,index / 2)*myPow(x,index / 2) * myPow(x,n-index);
    }

    public int getMaxTwo(int n){
        int index = 1;
        if (n / 2 > 0){
            index *= 2;
        }
        return index;
    }
}

• 这里忘记考虑负数的情况

public class Solution {
    /*
     * @param x: the base number
     * @param n: the power number
     * @return: the result
     */
    public double myPow(double x, int n) {
        // write your code here
        if (n == 0) return 1;
        if (n == 1) return x;
        if (n == -1) return 1/x;
        int index = getMaxTwo(n);
        return myPow(x,index / 2)*myPow(x,index / 2) * myPow(x,n-index);
    }

    public int getMaxTwo(int n){
        int index = 1;
        if (n / 2 != 0){
            index *= 2;
        }
        return n > 0 ? index : - index;
    }
}

• 在实际使用中，对于-Integer.MIN_VALUE这个值是不准确的，所以将其单独列出

switch (n) {
            case 0:
                return 1;
            case 1:
                return x;
            case -1:
                return 1/x;
            case Integer.MIN_VALUE:
                return 0;
        }

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• 首先暴力解决，时间复杂度过高（AC但时间较多），

public class Solution {
    /**
     * @param nums: the given array
     * @param k: the given number
     * @return:  whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k
     */
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        // Write your code here
        int length = nums.length;
        for (int i = 0; i < length ;i ++ ){
            for(int j = i + 1; j <= Math.min(length-1,i + k); j ++){
                if (nums[i] == nums[j])
                    return true;
            }
        } 
        return false;
    }
}

• 使用HashMap记录出现值的位置，判断是否在k以内（结果是内存爆炸）

public class Solution {
    /**
     * @param nums: the given array
     * @param k: the given number
     * @return:  whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k
     */
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        // Write your code here
        int length = nums.length;
        Map<Integer,Integer> map = new HashMap<Integer,Integer>();
        for (int i = 0; i < length ;i ++ ){
            if (map.get(nums[i]) == null){
                map.put(nums[i],i);
            } else {
                if (i - map.get(nums[i]) <= k){
                    return true;
                }
                map.put(nums[i],i);
            }
        } 
        return false;
    }
}

• 增加了超过K值以后删除，结果发现Map和Set性能无明显区别

public class Solution {
    /**
     * @param nums: the given array
     * @param k: the given number
     * @return:  whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k
     */
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        // Write your code here
        int length = nums.length;
        Map<Integer,Integer> map = new HashMap<Integer,Integer>();
        for (int i = 0; i < length ;i ++ ){
            if (map.get(nums[i]) == null){
                map.put(nums[i],i);
                if (i > k)
                    map.remove(nums[i - k]);
            } else {
                if (i - map.get(nums[i]) <= k){
                    return true;
                }
                map.put(nums[i],i);
            }
        } 
        return false;
    }
}

public boolean containsNearbyDuplicate(int[] nums, int k) {
        // Write your code here
        int length = nums.length;
        Set<Integer> map = new HashSet<Integer>();
        for (int i = 0; i < length ;i ++ ){
            if (!map.contains(nums[i])){
                map.add(nums[i]);
                if (i >= k)
                    map.remove(nums[i - k]);
            } else {
                return true;
            }
        } 
        return false;
    }

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• 排序之后直接遍历一次即可：

/**
 * Definition of Interval:
 * public classs Interval {
 *     int start, end;
 *     Interval(int start, int end) {
 *         this.start = start;
 *         this.end = end;
 *     }
 * }
 */

public class Solution {
    /**
     * @param intervals: interval list.
     * @return: A new interval list.
     */
    public List<Interval> merge(List<Interval> intervals) {
        // write your code here
        Collections.sort(intervals, new IntervalComparator()); 
        List<Interval> result = new ArrayList<>();
        Interval lastInterval = null;
        for (Interval interval : intervals){
            if (lastInterval == null || interval.start > lastInterval.end){
                lastInterval = interval;
                result.add(interval);
            } else {
                lastInterval.end = Math.max(lastInterval.end,interval.end);
            }
        } 
        return result;
    }

    private class IntervalComparator implements Comparator<Interval> {
        public int compare(Interval a, Interval b) {
            return a.start - b.start;
        }
    }
}

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git 地址： https://github.com/Outliwer/LintCode_Result