Problem

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.

Note: You may not slant the container and n is at least 2.

Thinking

自己思考的是用优化后的暴力解法,其中一个用例用了2ms,TLE了,可能时间限制比较严格。然后去Discuss大神区看了标准的解法,用的是双指针(哎,其实原先自己也有想到双指针,但差距就在于我不知道怎么设定双指针的移动条件,从而遍历出所有取得MaxArea的情况)。

大概思路:建立双支针,比较两指针所在的高度,移动高度较小位置的指针(如果移动高度较大位置的指针,由于指针向中间移动,所以下一个Area必定小于当前Area)


Answer

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public class ContainerWithMostWater {

// 正解,双指针,关键在于如何遍历maxArea的情况
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;
}


// 优化的暴力解法,TLE
// public int maxArea(int[] height) {
//
// int res = 0;
// int ix = 0;
// for (int i = 0; i < height.length-1; i++){
// if (height[i] < height[ix]) continue;
// for (int j = i+1; j < height.length; j++){
// int area = (j-i) * Math.min(height[i], height[j]);
// if (area >= res){
// res = area;
// ix = i;
// }
// }
// }
//
// return res;
// }
}