算法小记

只要在常数时间内可以将问题的大小削减为其一部分（$ \frac{1}{2} $）, 那么该算法就是($O(logN)$)

1. 最大子序列和问题($O(NlogN)$)

   public static int maxSubSum(int[] arr) {
   int maxSum = 0, thisSum = 0;
   for (int i = 0; i < arr.length; i++) {
       thisSum += arr[i];
       if (thisSum > maxSum) {
           maxSum = thisSum;
       } else if (thisSum < 0) {
           thisSum = 0;
       }
   }
   return maxSum;
   }

2. 折半查找binary search($O(logN)$)

   public static int binarySearch(int[] arr, int x) {
       int low = 0, high = arr.length - 1;
       while (low <= high) {
           int mid = (low + high)/2;
           if (arr[mid] < x) {
               low = ++mid;
           } else if (arr[mid] > x) {
               high = --mid;
           } else {
               return mid;
           }
       }
       return -1;
   }