leetcode day1

【所有详解可以参考】http://tianmaying.com/tutorials/tag/Leetcode?filter=hot&page=1

 

 

Two Sum:

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

解决思路：用一个map，for循环内，遍历数组中的加数1，map用来快速查询要找的加数2，如果找到，就返回map中的这个数，否则如果map没有此数，就put进去，key是数，value是数的index

public class Solution {
    public int[] twoSum(int[] numbers, int target) {

        HashMap<Integer,Integer> hash = new HashMap<Integer,Integer>();
        for(int i = 0; i < numbers.length; i++){

            Integer diff = (Integer)(target - numbers[i]);
            if(hash.containsKey(diff)){
                int toReturn[] = {hash.get(diff)+1, i+1};
                return toReturn;
            }

            hash.put(numbers[i], i);

        }

        return null;

    }
}

 

Remove Duplicates from Sorted Array:

Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.

thought：

用一个index洗刷掉数组中重复的元素，给数组重新赋值一遍，for循环遍历，当出现重复元素时，continue，否则，将下一个非重复的元素拿到前面来

public class Solution {
    public int removeDuplicates(int[] nums) {
    if(nums.length<=1){
        return nums.length;
        
    }
    int index = 0;
    for(int i=0; i<nums.length;i++){
        if(nums[index]==nums[i]){
           continue; 
        }else{
           nums[++index]=nums[i];
        }
    }
    return index+1;
    /*Set testSet = new HashSet();
    for(int i = 0; i<nums.length ; i++){
       testSet.add(new Integer(nums[i]));
    }
    return testSet.size();*/
    }
}