Container With Most Water
2017-08-05 11:07
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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.
/*
给定a[1],...,a
,求max((j-i)*min(a[i],a[j]));
从两边往中间扫
*/
class Solution {
public:
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;
}
};
..., 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.
/*
给定a[1],...,a
,求max((j-i)*min(a[i],a[j]));
从两边往中间扫
*/
class Solution {
public:
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;
}
};
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