1011. 在 D 天内送达包裹的能力

题解

二分法套路：
for (let i = 0; i < n; i++) {
if (isOK(i)) return ans;
}
连续的空间线性搜索，这就是二分查找可以发挥作用的标志。
const left = 最小可用；
const right = 最大可用;
while(left < right) {
const mid = Math.floor((left + right) / 2);
if(isOk(mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;

function shipWithinDays(weights: number[], D: number): number {
    let [left, right] = getMaxAndMin(weights);
    right += 1;
    while (left < right) {
        let mid = Math.floor((left + right) / 2);
        if (canFinish(weights, mid, D)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
};
function canFinish(weights: number[], weight: number, d: number) {
    let i: number = 0;
    for (let day = 0; day < d; day++) {
        let maxWeight = weight;
        while ((maxWeight -= weights[i]) >= 0) {
            // maxWeight -= weights[i];
            i++;
            if (i === weights.length) return true;
        }
    }
    return false;
}
function getMaxAndMin(weights: number[]) {
    let max = 0;
    let min = 0;
    for (let i of weights) {
        max += i;
        min = Math.max(min, i);
    }
    return [min, max];
}

同理题!!!!!

题解

function minEatingSpeed(piles: number[], h: number): number {
    let left = 1;
    let right = getMax(piles) + 1;
    while (left < right) {
        let mid = Math.floor((right + left) / 2);
        if (canFinish(piles, mid, h)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
};
function canFinish(piles, speed, h): boolean {
    let time = 0;
    for (let i = 0; i < piles.length; i++) {
        time += timeOf(piles[i], speed);
    }
    return time <= h;
}
function timeOf(n, speed) {
    return Math.floor(n / speed) + ((n % speed) > 0 ? 1 : 0);
}
function getMax(piles) {
    let max = 0;
    for (let i = 0; i < piles.length; i++) {
        max = Math.max(max, piles[i]);
    }
    return max;
}