每日一题6.24

Describe a O(nlgn)-time algorithm that,given a set S of n integers and another integer x,determines whether or not there exist two elements in S whose sum is exactly x.


类似于TwoSum类的问题,这里只是让返回是否存在二者之和恰好是x。只需要经过排序,然后采用对撞指针的方法来实现判断。

public static boolean TwoSum(int []arr,int target) throws Exception {
    if (arr==null||arr.length==0){
        throw new Exception("数组异常");
    }
    int i=1;
    int j=arr.length;
    MergeSort(arr,i-1,j-1);
    while (i<j){
        if (arr[i]+arr[j]==target){
            return true;
        }else if (arr[i]+arr[j]<target)
        {
            i++;
        }else {
            j--;
        }
    }
    return false;
}
public static void MergeSort(int []arr,int start,int end){
    int mid=start+(end-start)/2;
    MergeSort(arr,start,mid);
    MergeSort(arr,mid+1,end);
    merge(arr,start,mid,end);
}

private static void merge(int[] arr, int start, int mid, int end) {
    int [] help=new int[arr.length];
    int p=start;
    int q=mid+1;
    int i=0;
    while(p<=mid&&q<=end){
        help[i++]=arr[p]<arr[q]?arr[p++]:arr[q++];
    }
    if (p<=mid){
        help[i++]=arr[p++];
    }
    if (q<=end){
        help[i++]=arr[q++];
    }
    for (int j=0;j<arr.length;j++){
        arr[j]=help[j];
    }
}

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转载自blog.csdn.net/q_all_is_well/article/details/80855037