A[0] + A[1] + … + A[P−1] = A[P+1] + … + A[N−2] + A[N−1].

0 ≤ P < N

P 即为所求

数组中只有一个值可满足， 返回为0； 如找不到返回-1；

时间 空间复杂度都为o(n)

#include<stdio.h>

int solution(int A[], int N) {

    // write your code here…

    if (N<1)

        return -1;

    long long sum = 0;

    long long sumLeft =0;

    int i;

    for (i = 0; i < N; i++){

        sum += (long long)A[i];

    }   

    for (i = 0; i < N; i++){

        if (sumLeft == sum – sumLeft – A[i]){

            return i;

        }   

        sumLeft += (long long)A[i];

    }   

    return -1;

}

void main(){

    int A[7] = {-7, 1, 5, 2, -4, 3, 0};

    printf(“res = %d\n”, solution(A,7));

}

时间复杂度是o(n**2)

int solution(int A[], int N) {
    // write your code here...
    int i, l, k;
    long long sumPre;
    long long sumLast;
    for (i=0; i < N; i++){
      	sumPre =0;
    	sumLast =0;
        for (l = 0; l < i; l++){
        	sumPre += (long long)A[l];
        }
        for (l = i+1; l<N ; l++){
        	sumLast += (long long)A[l];
        }
        if (sumPre == sumLast)
            return i;    
    }
    return -1;
}