leetcode:11. Container With Most Water

説明
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.
構想
構想1:遍歴+減枝構想2:x軸距離最大の始まりに従い、徐々に下に縮小する
コード#コード#

public class Solution {
    public int maxArea(int[] height) {
        int maxArea = 0;
        int i = 0, j = height.length - 1;

        while(i < j){
            int bufArea = (height[i] > height[j] ? height[j] : height[i]) * (j - i);

            maxArea = maxArea>bufArea? maxArea:bufArea;

            if(height[i] < height[j])
                i++;
            else
                j--;
        }
        return maxArea;
    }
}

結果
他山の玉
C++ solutioin

int maxArea(vector<int>& height) {
    int water = 0;
    int i = 0, j = height.size() - 1;
    while (i < j) {
        int h = min(height[i], height[j]);
        water = max(water, (j - i) * h);
        while (height[i] <= h && i < j) i++;
        while (height[j] <= h && i < j) j--;
    }
    return water;
}

Java 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;
}