Container With Most Water(双指针)单调栈 leetcode11.
2018-03-08 00:08
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题目大意:给一个数组,其中数组在下标i处的值为A[i],坐标(i,A[i])和坐标(i,0)构成一条垂直于坐标轴x的直线。现任取两条垂线和x轴组成四边形容器。问其中盛水量最大为多少?
解法;从最原始的情况开始思考,即底的长度为最大值,两条竖直边分别是最左边和最右边,然后用left指向最左边,right指向最右边。然后考虑移动,在这种情况下,假如left>right,如果移动left的话,是没有意义的,根据短板原理,所盛水的体积取决于最小的那块板子,这种情况就应该移动较短的那块板子使得他变长。所以移动right向左移动,每次移动比较这次的体积与最大体积。直到right指向的板子比left的要长。即可
代码如下:int maxArea(vector<int>& height) {
int left = 0;
int right = height.size()-1;
int ma = 0,mi = 1e9;
while(left<right){
ma = max(ma, min(height[left], height[right])*(right-left));
mi = min(height[left],height[right]);
if(height[left]<height[right]){
left++;
}else{
right--;
}
}
return ma;
}
单调栈的坑先留在这里,肯定是可以用单调栈来做的,不过代码比较麻烦。
解法;从最原始的情况开始思考,即底的长度为最大值,两条竖直边分别是最左边和最右边,然后用left指向最左边,right指向最右边。然后考虑移动,在这种情况下,假如left>right,如果移动left的话,是没有意义的,根据短板原理,所盛水的体积取决于最小的那块板子,这种情况就应该移动较短的那块板子使得他变长。所以移动right向左移动,每次移动比较这次的体积与最大体积。直到right指向的板子比left的要长。即可
代码如下:int maxArea(vector<int>& height) {
int left = 0;
int right = height.size()-1;
int ma = 0,mi = 1e9;
while(left<right){
ma = max(ma, min(height[left], height[right])*(right-left));
mi = min(height[left],height[right]);
if(height[left]<height[right]){
left++;
}else{
right--;
}
}
return ma;
}
单调栈的坑先留在这里,肯定是可以用单调栈来做的,不过代码比较麻烦。
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