[leetcode]11. Container With Most Water

Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

分析：一般的思路是时间复杂度为O（n^2）的两层循环，选出最大的：

public int maxArea(int[] height) {
		int n = height.length;
		int res = 0;
		for(int i = 0; i < n-1; i++){
			for(int j = i+1; j < n; j++){
				int area = Math.min(height[i], height[j]) * (j - i);
				res = (area > res) ? area : res;
			}
		}
		return res;
    }

但是会超时，看到这样一个贪心的方法：

public class Solution {
    public int maxArea(int[] height) {
        int left = 0, right = height.length - 1;
    	int maxArea = 0;
    
    	while (left < right) {
    		maxArea = Math.max(maxArea, Math.min(height[left], height[right])
    				* (right - left));
    		if (height[left] < height[right])
    			left++;
    		else
    			right--;
    	}
    
    	return maxArea;
    }
}

贪心的证明：假如A[0] < A[n - 1]，那么对于任意k(k < n - 1)，A[0]到A[k]能盛的水一定小于A[0]到A[n - 1]的盛水，所以我们只要关注A[1]到A[n - 1]的最大盛水量就好了。综上，每次内移小的一方，同时跟踪记录最大值。