• 1. 数组之差TapeEquilibrium Minimize the value |(A[0] + ... + A[P-1])


    数组之差

    package com.code;
    
    public class Test03_3 {
         public static int solution(int[] A) {
             int size = A.length;
             if (size<2){
                 return -1;
             }
             int [] rightSum = new int[size];
             rightSum[size-1] = A[size-1];
             for(int i=size-2;i>=0;i--){
                 rightSum[i] = A[i]+rightSum[i+1];
             }
             int [] leftSum = new int[size];
             leftSum[0] = A[0];
             for(int i=1;i<size-1;i++){
                 leftSum[i] = A[i]+leftSum[i-1];
             }
             int min = 2147483647;
             for(int i=0;i<size-1;i++){
                 min = Math.min(min, Math.abs(leftSum[i]-rightSum[i+1]));
             }
             return min;
         }
        public static void main(String[] args) {
            int a [] = {3,1,2,4,3};
            System.out.println(solution(a));
            int b[] = {1,3};
            System.out.println(solution(b));
        }
    }
    
    /**
    
    A non-empty zero-indexed array A consisting of N integers is given. Array A represents numbers on a tape.
    
    Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
    
    The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|
    
    In other words, it is the absolute difference between the sum of the first part and the sum of the second part.
    
    For example, consider array A such that:
    
      A[0] = 3
      A[1] = 1
      A[2] = 2
      A[3] = 4
      A[4] = 3
    We can split this tape in four places:
    
    P = 1, difference = |3 − 10| = 7 
    P = 2, difference = |4 − 9| = 5 
    P = 3, difference = |6 − 7| = 1 
    P = 4, difference = |10 − 3| = 7 
    Write a function:
    
    class Solution { public int solution(int[] A); }
    
    that, given a non-empty zero-indexed array A of N integers, returns the minimal difference that can be achieved.
    
    For example, given:
    
      A[0] = 3
      A[1] = 1
      A[2] = 2
      A[3] = 4
      A[4] = 3
    the function should return 1, as explained above.
    
    Assume that:
    
    N is an integer within the range [2..100,000];
    each element of array A is an integer within the range [−1,000..1,000].
    Complexity:
    
    expected worst-case time complexity is O(N);
    expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
    Elements of input arrays can be modified.
    */
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  • 原文地址:https://www.cnblogs.com/stono/p/6419148.html
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