强化四 二分答案

lt75 Find Peak Element
lt390 Find Peak Element II
lt141 Sqrt(x)
lt586 Sqrt(x) II 注意判断x与1的大小关系
二分答案 步骤：确定答案范围 找一个验证答案的方法。 时间复杂度O(log(n)*(验证一个答案的事件))
lt183 Wood Cut 答案范围 1 - maxlength；验证函数 是否能切大于等于k段
lt437 Copy Books 答案范围 最大页数 - 全部页数和；验证函数 是否能少于等于k个人完成
lt633 Find the Duplicate Number 答案范围 1 - n； 验证函数 小于等于该元素的元素个数是否小于等于该元素本身

lt75 Find Peak Element O(logn)

public class Solution {
    /**
     * @param A: An integers array.
     * @return: return any of peek positions.
     */
    public int findPeak(int[] A) {
        // write your code here
        int left = 0;
        int right = A.length-1;
        while(left+1A[mid+1] && A[mid]>A[mid-1])
                return mid;
            if(A[mid]>A[mid+1] && A[mid]A[right] ? left : right;
    }
}

lt390 Find Peak Element II

下面给出的是O(nlogn)解法
还能更快到O(n) 横竖横竖切割

public class Solution {
    /*
     * @param A: An integer matrix
     * @return: The index of the peak
     */
    public List findPeakII(int[][] A) {
        // write your code here
        List result = new ArrayList<>();
        if(A==null || A.length==0 || A[0]==null || A[0].length==0)
            return result;
            
        for(int i=1; iA[i-1][index] && A[i][index]>A[i+1][index]){
                result.add(i);
                result.add(index);
                return result;
            }
        }
        return result;
    }
    
    public int findPeak(int[] A) {
        // write your code here
        int left = 0;
        int right = A.length-1;
        while(left+1A[mid+1] && A[mid]>A[mid-1])
                return mid;
            if(A[mid]>A[mid+1] && A[mid]A[right] ? left : right;
    }
}

lt141 Sqrt(x)

public class Solution {
    /**
     * @param x: An integer
     * @return: The sqrt of x
     */
    public int sqrt(int x) {
        // write your code here
        int left = 0;
        int right = x;
        while(left+1x/mid){
                right = mid;
            }else if(mid

lt586 Sqrt(x) II

public class Solution {
    /**
     * @param x: a double
     * @return: the square root of x
     */
    public double sqrt(double x) {
        // write your code here
        double left = 0;
        double right = Math.max(1.0, x);
        double eps = 1e-12;
        while(left+epsx/mid){
                right = mid;
            }else if(mid

lt183 Wood Cut 答案范围 1 - maxlength

public class Solution {
    /**
     * @param L: Given n pieces of wood with length L[i]
     * @param k: An integer
     * @return: The maximum length of the small pieces
     */
    public int woodCut(int[] L, int k) {
        // write your code here
        if(L==null || L.length==0 || k<=0)
            return 0;
        int max = findMax(L);
        int left = 1;
        int right = max;
        while(left+1=k){
                left = mid;
            }else{
                right = mid;
            }
        }
        if(valid(right, L, k) >= k)
            return right;
        if(valid(left, L, k) >= k)
            return left;
        return 0;
    }
    private int findMax(int[] L){
        int max = L[0];
        for(int i=0; i

lt437 Copy Books 答案范围 最大页数 - 全部页数和

public class Solution {
    /**
     * @param pages: an array of integers
     * @param k: An integer
     * @return: an integer
     */
    public int copyBooks(int[] pages, int k) {
        // write your code here
        
        int min = findMax(pages);
        int max = getSum(pages);
        while(min+1k){
                min = mid;
            }else{
                max = mid;
            }
        }
        if(peopleNeed(pages, min)<=k)
            return min;
        return max;
    }
    private int peopleNeed(int[] pages, int hour){
        int count = 0;
        int sum = 0;
        for(int i=0; ihour){
                sum = pages[i];
                count++;
            }else if(sum+pages[i]==hour){
                sum = 0;
                count++;
            }else{
                sum = sum+pages[i];
            }
        }
        if(sum>0)
            count++;
        return count;
    }
    private int findMax(int[] pages){
        int max = 0;
        for(int i: pages){
            max = Math.max(max, i);
        }
        return max;
    }
    private int getSum(int[] pages){
        int sum = 0;
        for(int i: pages){
            sum = sum + i;
        }
        return sum;
    }
}

lt633 Find the Duplicate Number

public class Solution {
    /**
     * @param nums: an array containing n + 1 integers which is between 1 and n
     * @return: the duplicate one
     */
    public int findDuplicate(int[] nums) {
        // write your code here
        if(nums==null || nums.length==0)
            return -1;
        int n = nums.length-1;
        int left = 1;
        int right = n;
        while(left+1left)
            return left;
        return right;
    }
    private int lessOrEqual(int num, int[] nums){
        int count = 0;
        for(int i: nums){
            if(i<=num)
                count++;
        }
        return count;
    }
}