【LeetCode】【Python题解】Container with most water
2014-10-08 21:30
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这道题真真是在面试中碰到过,可惜当时复杂度到O(n2)了,太挫了,怪不得没有通过面试。
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.
题目说的有点复杂,大意是利用x轴作底,两个任意的竖直线段作杯壁,何时盛水最多。
木桶原理大家肯定都知道,水盛多少取决于最短的杯壁,所以此题还可以引申为往围成的区域内放矩形,怎样使得矩形面积最大。题目中的不能倾斜(slant:倾斜,倾倒)对应为矩形必须水平放置。
复杂度为O(n)的思想是贪心原理,先从底边最大的情况考虑,计算最大面积后,此时要将底边长度减1,只需要将杯壁较短的那一边移动一个单位距离,得到的解必定优于杯壁较长那边移动的情况。这样保证每次移动都得到的是局部最优解。
class Solution {
public:
int maxArea(vector<int> &height) {
int Len = height.size(),low=0,high=Len-1;
int maxV = 0;
while(low<high){
maxV = max(maxV,(high-low)*min(height[low],height[high]));
if (height[low]<height[high]) low++;
else high--;
}
return maxV;
}
};
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.
题目说的有点复杂,大意是利用x轴作底,两个任意的竖直线段作杯壁,何时盛水最多。
木桶原理大家肯定都知道,水盛多少取决于最短的杯壁,所以此题还可以引申为往围成的区域内放矩形,怎样使得矩形面积最大。题目中的不能倾斜(slant:倾斜,倾倒)对应为矩形必须水平放置。
复杂度为O(n)的思想是贪心原理,先从底边最大的情况考虑,计算最大面积后,此时要将底边长度减1,只需要将杯壁较短的那一边移动一个单位距离,得到的解必定优于杯壁较长那边移动的情况。这样保证每次移动都得到的是局部最优解。
class Solution {
public:
int maxArea(vector<int> &height) {
int Len = height.size(),low=0,high=Len-1;
int maxV = 0;
while(low<high){
maxV = max(maxV,(high-low)*min(height[low],height[high]));
if (height[low]<height[high]) low++;
else high--;
}
return maxV;
}
};
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