2017.10.19 LeetCode 二分 -> 69. Sqrt(x) -> 410. Split Array Largest Sum

69. Sqrt(x)

Description

Implement int sqrt(int x).

Compute and return the square root of x.

题意：让你求x√

分析： 直接二分即可，注意好边界条件即可

class Solution {
public:
bool check(int x,int mid){
if(mid*1ll*mid <= 1ll*x){
return true;
}
return false;
}

int mySqrt(int x) {
int l = 0,r = x,ans,mid;
while(l <= r){
mid = l + r >> 1;
if(check(x,mid)){
ans = mid;
l = mid + 1;
}
else {
r = mid - 1;
}
}
return ans;
}
};

410. Split Array Largest Sum

Description

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.

Note:

If n is the length of array, assume the following constraints are satisfied:

1 ≤ n ≤ 1000

1 ≤ m ≤ min(50, n)

Examples:

Input:

nums = [7,2,5,10,8]

m = 2

Output:

18

Explanation:

There are four ways to split nums into two subarrays.

The best way is to split it into [7,2,5] and [10,8],

where the largest sum among the two subarrays is only 18.

题意： 给你一个数组，和一个m，让你把这个数组分成连续的m块（非空），让你求每一块的和的最小值（有点拗口）范围：1≤n≤1000 1≤m≤min(50,n)

分析：先把题意搞懂把，什么叫每一块和的最小值呢，简单的说就是分好后，计算每一块的sum，然后在所有块中取一个最大的sum，就是题目要找的，看似毫无头绪的题，细心发现其实是有规律的，大家都知道二分呢只要满足单调性和可判定性就可以使用啦，我们先找单调性，先看样例上的答案 18，为什么17不行呢，先带入下，如果答案是17的话，先是前三个[7, 2, 5]，然后再是[10],再是[8],我们发现他是分了三块了，而要求为两块，显然不符合，这就是我们要找的单调性，我们发现，我们随着这个答案的变大，分的块数是变少的，这里呢单调性搞定了，在找判定性，也很好找，以为题目上给了m，直接比较即可，这时我们就可以放心大胆地使用二分了，当然啦还得注意一些地方，比如：是否会爆int，边界问题，等等，这些自己体会吧.

参考函数

class Solution {
public:
bool check(vector<int> &nums,long long mid,int m) {
long long sum = 0;
int len = nums.size(),t = 0;
for(int i = 0;i < len;i++) {
if(sum + nums[i] > mid) {
t++;
sum = nums[i];
if(sum > mid) return false;
}else {
sum += nums[i];
}
}
if(t+1 <= m)
return true;
return false;
}

int splitArray(vector<int>& nums, int m) {
long long sum = 0;
int len = nums.size();
for(int i = 0;i < len;i++) {
sum += 1ll*nums[i];
}
long long  l = 0,r = sum,mid,ans;
while(l <= r){
mid = r + l >>1;
if(check(nums,mid,m)) {
ans = mid;
r = mid-1;
}else {
l = mid+1;
}
}
return (int)ans;
}
};