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;
	}